Vectors Exercise 01
Question:
Area of a rectangle having vertices A, B, C and D with position vectors −i^+1/2j^+4k^, i^+1/2j^+4k^, i^−1/2j^+4k^ and −i^−1/2j^+4k^ respectively is A 21 B 1 C 2 D 4
Answer:
Answer: D 4
Question:
Show that the points A(1,2,7),B(2,6,3) and C(3,10,−1) are collinear .
Answer:
Step 1: Calculate the vector AB. AB = (21, 62, 37) = (1, 4, 4)
Step 2: Calculate the vector AC. AC = (31, 102, 17) = (2, 8, 8)
Step 3: Calculate the cross product of the vectors AB and AC. AB x AC = (1 x 8  4 x 2, 4 x 2  1 x 8, 4 x 8  (4) x 2) = (6, 12, 32)
Step 4: If the cross product is equal to zero, then the points are collinear. Since the cross product of AB and AC is not equal to zero, the points A, B, and C are not collinear.
Question:
If a unit vector a makes an angle π/3 with i^,π/4 with j^ and an acute angle θ with k^, then find θ and hence, the components of a.
Answer:

Since a is a unit vector, its magnitude is 1.

Using the dot product, we can find the angle θ between a and k^.
a•k^ = ak^cosθ
1•1cosθ = cosθ
θ = cos1(1)
θ = 0°
 Now we can find the components of a.
a = (cosπ/3, cosπ/4, cos0°)
a = (1/2, √2/2, 1)
Question:
Find the values of x and y so that the vectors 2i^+3j^ and xi^+yj^ are equal
Answer:
Step 1: Set the two vectors equal to each other: 2i^+3j^ = xi^+yj^
Step 2: Subtract xi^ from both sides of the equation: 2i^+3j^  xi^ = xi^+yj^  xi^
Step 3: Simplify the left side of the equation: 2i^  xi^ + 3j^ = yj^
Step 4: Isolate the xi^ term on the left side of the equation: xi^ = 2i^  3j^
Step 5: Substitute the expression for xi^ into the original equation: 2i^+3j^ = (2i^  3j^) + yj^
Step 6: Simplify the left side of the equation: 2i^ + 3j^ = 2i^  3j^ + yj^
Step 7: Isolate the yj^ term on the left side of the equation: yj^ = 3j^
Step 8: Substitute the expression for yj^ into the original equation: 2i^+3j^ = xi^ + (3j^)
Step 9: Substitute the expression for xi^ into the equation: 2i^+3j^ = (2i^  3j^) + (3j^)
Step 10: Simplify and solve the equation: 2i^+3j^ = 2i^ + 0j^
Step 11: The final solution is x = 2 and y = 3.
Question:
Answer the following as true or false. (i) a and −a are collinear (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal.
Answer:
(i) False (ii) False (iii) False (iv) True
Question:
Show that the vectors 2i^−j^+k^,i^−3j^−5k^ and 3i^−4j^−4k^ form the vertices of a right angled triangle .
Answer:
Step 1: Calculate the dot product of the vectors: 2i^−j^+k^⋅i^−3j^−5k^ = 2i^2−j^2−5k^2 = 2−1−5 = −4
Step 2: Calculate the dot product of the vectors: 2i^−j^+k^⋅3i^−4j^−4k^ = 6i^2−4j^2−4k^2 = 6−4−4 = −2
Step 3: Calculate the dot product of the vectors: i^−3j^−5k^⋅3i^−4j^−4k^ = 3i^2−4j^2−4k^2 = 3−4−4 = −5
Step 4: Calculate the length of the vectors using the Pythagorean theorem: a^2+b^2=c^2
2i^−j^+k^: (2)^2+(−1)^2+(1)^2 = 4+1+1 = 6
i^−3j^−5k^: (1)^2+(−3)^2+(−5)^2 = 1+9+25 = 35
3i^−4j^−4k^: (3)^2+(−4)^2+(−4)^2 = 9+16+16 = 41
Step 5: Verify that the dot product of any two vectors is equal to the product of their lengths minus the length of the third vector: −4 = 6⋅35−41 −2 = 6⋅41−35 −5 = 35⋅41−6
Step 6: Therefore, the vectors 2i^−j^+k^,i^−3j^−5k^ and 3i^−4j^−4k^ form the vertices of a right angled triangle.
Question:
Find ∣a×b∣, if a=i^−7j^+7k^ and b=3i^−2j^+2k^ .
Answer:
Answer: Step 1: Multiply a and b a×b = (i^−7j^+7k^) × (3i^−2j^+2k^)
Step 2: Simplify a×b = 3i^−9j^+14k^
Step 3: Find the absolute value of a×b ∣a×b∣ = √(3^2 + (9)^2 + 14^2)
Step 4: Solve ∣a×b∣ = √(9 + 81 + 196)
Step 5: Simplify ∣a×b∣ = √286
Therefore, ∣a×b∣ = 17
Question:
Show that the points A,B and C with position vectors, a=3i^−4j^−4k^,b=2i^−j^+k^ and c=i^−3j^−5k^, respectively form the vertices of a right angled triangle.
Answer:

Find the magnitude of each of the three vectors a,b and c: a = √(3^2 + (4)^2 + (4)^2) = 5 b = √(2^2 + (1)^2 + (1)^2) = √4 = 2 c = √(1^2 + (3)^2 + (5)^2) = √34 = 5.83

Find the magnitude of the vector AB: AB = √[(23)^2 + (1+4)^2 + (45)^2] = √50 = 7.07

Find the magnitude of the vector AC: AC = √[(13)^2 + (34)^2 + (54)^2] = √50 = 7.07

Find the magnitude of the vector BC: BC = √[(12)^2 + (3+1)^2 + (51)^2] = √26 = 5.1

Use the Pythagorean theorem to check if the triangle is right angled: a^2 + b^2 = c^2 5^2 + 2^2 = 5.83^2 25 + 4 = 34.69 29 = 34.69
Since 29 ≠ 34.69, the points A,B and C do not form the vertices of a right angled triangle.
Question:
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (−5,7)
Answer:
Scalar component: The scalar component is the magnitude of the vector, which is calculated using the Pythagorean theorem.
√((5  2)^2 + (7  1)^2) = √(49 + 36) = √85 = 9.22
Vector component: The vector component is the direction of the vector, which is calculated using the inverse tangent.
tan1((7  1)/(5  2)) = tan1(6/7) = 0.927
Question:
Represent graphically a displacement of 40 km, 30o east of north.
Answer:
Answer:
Step 1: Draw the xaxis and yaxis on the graph paper.
Step 2: Mark the origin (0,0) on the graph paper.
Step 3: Draw a line from the origin (0,0) to the point (40,0). This line represents the displacement of 40 km.
Step 4: Rotate the line 30o clockwise around the origin.
Step 5: The point at the end of the line represents the displacement of 40 km, 30o east of north.
Question:
Let the vectors a,b,c given as a1i^+a2j^+a3k^,b1i^+b2j^+b3k^,c1i^+c2j^+c3k^. Then show that =a×(b+c)=a×b+a×c.
Answer:
Given : a = a1i + a2j + a3k b = b1i + b2j + b3k c = c1i + c2j + c3k
To show that a × (b + c) = a × b + a × c
Step 1: Calculate a × (b + c)
a × (b + c) = (a1i + a2j + a3k) × (b1i + b2j + b3k + c1i + c2j + c3k)
Step 2: Expand the equation
a × (b + c) = (a1i + a2j + a3k) × (b1i + b2j + b3k) + (a1i + a2j + a3k) × (c1i + c2j + c3k)
Step 3: Use the distributive property
a × (b + c) = a1i × (b1i + b2j + b3k) + a2j × (b1i + b2j + b3k) + a3k × (b1i + b2j + b3k) + a1i × (c1i + c2j + c3k) + a2j × (c1i + c2j + c3k) + a3k × (c1i + c2j + c3k)
Step 4: Simplify the equation
a × (b + c) = a1i × b1i + a1i × b2j + a1i × b3k + a2j × b1i + a2j × b2j + a2j × b3k + a3k × b1i + a3k × b2j + a3k × b3k + a1i × c1i + a1i × c2j + a1i × c3k + a2j × c1i + a2j × c2j + a2j × c3k + a3k × c1i + a3k × c2j + a3k × c3k
Step 5: Rearrange the equation
a × (b + c) = (a1i × b1i + a2j × b2j + a3k × b3k) + (a1i × b2j + a2j × b1i + a1i × b3k + a3k × b1i + a2j × b3k + a3k × b2j) + (a1i × c1i + a2j × c2j + a3k × c3k) + (a1i × c2j + a2j × c1i + a1i × c3k + a3k × c1i + a2j × c3k + a3k × c2j)
Step 6: Use the vector product property
a × (b + c) = a × b + a × c
Question:
Evaluate the product (3a−5b)⋅(2a+7b) .
Answer:
 (3a−5b)⋅(2a+7b)
 3a⋅2a + 3a⋅7b  5b⋅2a  5b⋅7b
 6a² + 21ab  10ab  35b²
 6a² + 11ab  35b²
Question:
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i^+2j^−k^ and −i^+j^+k^ respectively, in the ratio 2:1, (i) internally (ii) externally
Answer:
(i) Internally:
Let the position vector of the point R be ai^+bj^+ck^
According to the given condition,
PR:QR = 2:1
⇒ PR = 2QR
⇒ ai^+bj^+ck^  (−i^+j^+k^) = 2(i^+2j^−k^)
⇒ (a+1)i^+(b+2)j^+(c−1)k^ = 2i^+4j^−2k^
Comparing the coefficients of i^, j^ and k^,
a+1 = 2, b+2 = 4, c−1 = −2
⇒ a = 1, b = 2, c = −1
Therefore, the position vector of the point R which divides the line joining two points P and Q internally in the ratio 2:1 is i^+2j^−k^.
(ii) Externally:
Let the position vector of the point R be ai^+bj^+ck^
According to the given condition,
PR:QR = 2:1
⇒ PR = 2QR
⇒ ai^+bj^+ck^  (i^+2j^−k^) = 2(−i^+j^+k^)
⇒ (a−1)i^+(b−2)j^+(c+1)k^ = −2i^+2j^+2k^
Comparing the coefficients of i^, j^ and k^,
a−1 = −2, b−2 = 2, c+1 = 2
⇒ a = 1, b = 0, c = 1
Therefore, the position vector of the point R which divides the line joining two points P and Q externally in the ratio 2:1 is i^+k^.
Question:
Answer the following as true or false. a and −a are collinear A True B False C Ambiguous D Data insufficient
Answer:
B False
Question:
If the vertices A,B,C of a triangle ABC are (1,2,3),(−1,0,0),(0,1,2), respectively then find ∠ABC. [∠ABC is the angle between the vectors BA and BC] .
Answer:

Find the vectors BA and BC. BA = (1 (1), 20, 30) = (2,2,3) BC = (01, 10, 23) = (1,1,1)

Find the magnitude of vectors BA and BC. BA = √(2² + 2² + 3²) = √(13) BC = √(1² + 1² + 1²) = √(3)

Find the dot product of the vectors BA and BC. BA . BC = (2*1) + (2*1) + (3*1) = 8

Find the angle between the vectors BA and BC. ∠ABC = cos⁻¹(8/√(13)*√(3)) = cos⁻¹(8/13) = cos⁻¹(0.615384615) = 115.9°
Question:
Find the angle between two vectors a and b with magnitudes √3 and 2, respectively having a⋅b=√6 Answer required
Answer:

First, calculate the dot product of the two vectors, a and b. This is done by multiplying the two vectors together, element by element. a⋅b = a1b1 + a2b2 + a3b3 = √3*2 = √6

Next, calculate the magnitudes of the two vectors. The magnitude of a vector is the length of the vector, which can be found using the Pythagorean theorem. The magnitude of vector a is √3 and the magnitude of vector b is 2.

Finally, use the formula for the angle between two vectors, which is: θ = cos1(a⋅b/a*b)

Plugging in the values for a⋅b and ab, we get: θ = cos1(√6/(√32)) = cos1(1/3) = 60°
Therefore, the angle between the two vectors is 60°.
Question:
Let the vectors a and b be such that ∣a∣=3 and ∣b∣=√2/3, then a×bis a unit vector, if the angle between aand b is A π/6 B π/4 C π/3 D π/2
Answer:
Answer: B π/4
Question:
Show that the vector i^+j^+k^ is equally inclined to the axes OX,OY and OZ.
Answer:
Step 1: A vector is said to be equally inclined to the axes OX,OY and OZ if its components along the axes OX,OY and OZ are equal.
Step 2: The vector i^+j^+k^ has components along the axes OX,OY and OZ as 1,1 and 1 respectively.
Step 3: Since the components of the vector i^+j^+k^ along the axes OX,OY and OZ are equal, the vector is equally inclined to the axes OX,OY and OZ.
Question:
If a and b are two collinear vectors, then which of the following are incorrect? A b =λa , for some scalar λ B a =±b C the respective components of a and b are proportional D both the vectors a and b have same direction, but different magnitudes
Answer:
A. a = λb, for some scalar λ is correct B. a = ±b is incorrect C. the respective components of a and b are proportional is correct D. both the vectors a and b have same direction, but different magnitudes is correct
Question:
Find λ and μ if (2i^+6j^+27k^)×(i^+λj^+μk^)=0
Answer:

(2i^+6j^+27k^)×(i^+λj^+μk^) = 0

2i^+6j^+27k^+i^λj^+μk^ = 0

i^(2+λ) + j^(6+μ) + k^(27+μ) = 0

i^(2+λ) = 0, j^(6+μ) = 0, k^(27+μ) = 0

2+λ = 0, 6+μ = 0, 27+μ = 0

λ = 2, μ = 6
Question:
Find the direction cosines of the vector joining the points A(1,2,−3) and B(−1,−2,1) directed from A to B.
Answer:
Step 1: Find the vector joining the points A and B.
Vector AB = (A  B) = (1, 2, 3)  (1, 2, 1) = (2, 4, 4)
Step 2: Find the magnitude of the vector AB.
AB = √(2^2 + 4^2 + (4)^2) = √36 = 6
Step 3: Find the direction cosines of the vector AB.
cos α = 2/6, cos β = 4/6, cos γ = 4/6
Question:
Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 metres northwest (iii) 40o (iv) 40 watt (v) 10−19 coluomb (vi) 20 m/s^2
Answer:
(i) 10 kg  Scalar (ii) 2 metres northwest  Vector (iii) 40o  Scalar (iv) 40 watt  Scalar (v) 10−19 coluomb  Scalar (vi) 20 m/s^2  Vector
Question:
Find the unit vector in the direction of vector PQ, where P and Q are the points (1,2,3) and (4,5,6), respectively Answer required
Answer:
Step 1: Find the vector PQ by subtracting the coordinates of P from the coordinates of Q. PQ = (4,5,6)  (1,2,3) = (3,3,3)
Step 2: Calculate the magnitude of the vector PQ. Magnitude of PQ = √(3^2 + 3^2 + 3^2) = √27 = 3√3
Step 3: Divide the components of the vector PQ by its magnitude to find the unit vector in the direction of PQ. Unit vector in the direction of PQ = (3/3√3, 3/3√3, 3/3√3) = (1/√3, 1/√3, 1/√3)
Question:
If either a =0 or b =0 , then a ×b =0 . Is the converse true? Justify your answer with an example .
Answer:
Answer: No, the converse is not true.
For example, if a = 2 and b = 0, then a × b = 0. However, if a × b = 0, it does not necessarily mean that either a = 0 or b = 0. For example, if a = 0 and b = 2, then a × b = 0.
Question:
Given that a ⋅b =0 and a ×b =0 . What can you conclude about the vectors a and b ?
Answer:

Since a⋅b = 0, the vectors a and b are perpendicular to each other.

Since a×b = 0, the vectors a and b have either the same direction or opposite directions.
Question:
Show that (a −b )×(a +b )=2(a ×b)
Answer:
(a \n −b \n )×(a \n +b \n )
= (a \n ×a \n )  (a \n ×b \n )  (b \n ×a \n ) + (b \n ×b \n )
= a^2  a \n ×b \n  b \n ×a \n + b^2
= a^2  a \n ×b \n  a \n ×b \n + b^2
= a^2  2a \n ×b \n + b^2
= 2(a \n ×b \n )
= 2(a \n ×b)
Question:
Find the projection of the vector i^+3j^+7k^ on the vector 7i^−j^+8k^ .
Answer:
Answer: Step 1: Find the magnitude of the vector 7i^−j^+8k^ Magnitude = √(7^2 + (1)^2 + 8^2) = √(50) = 7.07
Step 2: Find the dot product of the vectors i^+3j^+7k^ and 7i^−j^+8k^ Dot product = (i^+3j^+7k^) • (7i^−j^+8k^) = (17) + (3(1)) + (7*8) = 59
Step 3: Find the projection of the vector i^+3j^+7k^ on the vector 7i^−j^+8k^ Projection = (59/50) * (7i^−j^+8k^) = (59/50) * (7i^−j^+8k^) = 6.12i^−0.76j^+9.76k^
Question:
Show that ∣a ∣b +∣b ∣a is perpendicular to ∣a ∣b −∣b ∣a , for any two nonzero vectors a and b .
Answer:
 First, let’s define the two vectors:
∣a \n ∣b \n +∣b \n ∣a \n = v_1 = (a_1, b_1, b_1, a_1)
∣a \n ∣b \n −∣b \n ∣a \n = v_2 = (a_2, b_2, b_2, a_2)
 Now, we can use the dot product to determine if the two vectors are perpendicular:
v_1 · v_2 = a_1a_2 + b_1b_2 + b_1(b_2) + a_1(a_2)
 Simplifying, we get:
v_1 · v_2 = a_1a_2 + b_1b_2  b_1b_2  a_1a_2
 Finally, we can see that the dot product is equal to 0, which tells us that the two vectors are perpendicular:
v_1 · v_2 = 0
Question:
Find a vector parallel to the given vector 5i^−j^+2k^ which has magnitude 8 units.
Answer:
Step 1: Find the unit vector of the given vector 5i^−j^+2k^.
Unit vector = (5/√29)i^−(1/√29)j^+(2/√29)k^
Step 2: Multiply the unit vector by 8 to find the vector parallel to the given vector with magnitude 8 units.
Vector = (40/√29)i^−(8/√29)j^+(16/√29)k^
Question:
If either vector a =0 or b =0 , then a ⋅b =0. But the converse need not be true. Justify your answer with an example.
Answer:
Answer:

The statement is true because if either vector a or b is equal to 0, then the dot product of a and b will be equal to 0.

To illustrate, let vector a = (1, 0) and vector b = (0, 2). The dot product of a and b is 0, but neither vector a or vector b is equal to 0. Therefore, the converse need not be true.
Question:
If a ⋅a =0 and a ⋅b =0, then what can be concluded about the vector b ?
Answer:

Since a ⋅a = 0, we can conclude that the vector a is a zero vector.

Since a ⋅b = 0, we can conclude that the vector b is either a zero vector or is perpendicular to vector a.
Question:
Find the direction cosines of the vector i^+2j^+3k^.
Answer:
Answer: Step 1: Find the magnitude of the vector i^+2j^+3k^.
Magnitude = √(1^2 + 2^2 + 3^2)
Magnitude = √14
Step 2: Find the direction cosines of the vector i^+2j^+3k^.
Direction cosines = (1/√14 , 2/√14 , 3/√14)
Answer: The direction cosines of the vector i^+2j^+3k^ are (1/√14 , 2/√14 , 3/√14).
Question:
If a =2i^+2j^+3k^,b =−i^+2j^+k^ and c =3i^+j^ are such that a +λb is a perpendicular to c , then find the value of λ.
Answer:

Write down the vector equations for a and b: a = 2i + 2j + 3k b = i + 2j + k

Write down the vector equation for c: c = 3i + j

To find the value of λ, we need to find the dot product of vectors a and b: a·b = (2i + 2j + 3k)·(i + 2j + k)

Simplify the dot product equation: a·b = 2i·(i) + 2j·2j + 3k·k

Substitute the values of the components of the vectors: a·b = 2 + 4 + 3

Calculate the dot product: a·b = 5

To find the value of λ, we need to find the cross product of vectors a and c: a x c = (2i + 2j + 3k) x (3i + j)

Simplify the cross product equation: a x c = (2i x 3i) + (2j x j) + (3k x 0)

Substitute the values of the components of the vectors: a x c = 6i  2j

Calculate the cross product: a x c = 6i  2j

To find the value of λ, we need to take the dot product of the cross product of a and c with b: (a x c)·b = (6i  2j)·(i + 2j + k)

Simplify the dot product equation: (a x c)·b = 6i·(i)  2j·2j + (2j + k)·k

Substitute the values of the components of the vectors: (a x c)·b = 6 + 4 + 2

Calculate the dot product: (a x c)·b = 2

Divide the dot product of a and b by the dot product of the cross product of a and c with b: λ = 5/2

Therefore, the value of λ is 2.5.
Question:
Find the area of the triangle with vertices A(1,1,2),B(2,3,5) and C(1,5,5) .
Answer:
Step 1: Calculate the length of the sides of the triangle using the distance formula.
AB = √((21)2 + (31)2 + (52)2) = √(12 + 4 + 9) = √25 = 5
BC = √((12)2 + (53)2 + (55)2) = √(1 + 4 + 0) = √5 = 2.24
AC = √((11)2 + (51)2 + (52)2) = √(0 + 16 + 9) = √25 = 5
Step 2: Calculate the semiperimeter (s) of the triangle.
s = (5 + 2.24 + 5)/2 = 12.24/2 = 6.12
Step 3: Calculate the area of the triangle using Heron’s formula.
Area = √(6.12(6.12  5)(6.12  2.24)(6.12  5)) = √(6.12(1.12)(3.88)(1.12)) = √10.14 = 3.18
Question:
Classify the following as scalar and vector quantities. (i) Time period (ii) Distance (iii) Force (iv) Velocity (v) Work done
Answer:
(i) Time period  Scalar (ii) Distance  Scalar (iii) Force  Vector (iv) Velocity  Vector (v) Work done  Scalar
Question:
Two vectors having same magnitude are collinear A True B False C Ambiguous D Data insufficient
Answer:
A: False B: False C: Ambiguous D: Data insufficient
Question:
Find the sum of the vectors a =i^−2j^+k^,b =−2i^+4j^+5k^ and c =i^−6j^−7k^.
Answer:
a) Find the components of vector a: a = i^2j^+k^
b) Find the components of vector b: b = 2i^+4j^+5k^
c) Find the components of vector c: c = i^6j^7k^
d) Find the sum of the vectors a, b and c: a + b + c = (i^2j^+k^) + (2i^+4j^+5k^) + (i^6j^7k^) = (i^+2j^+6k^)
Question:
Find the magnitude of two vectors a and b, having the same magnitude and such that the angle between them is 60∘ and their scalar product is 21 .
Answer:
Step 1: Calculate the magnitude of the vectors a and b.
The magnitude of a vector is given by the formula a = √(a1^2 + a2^2 + … + an^2).
Since the two vectors have the same magnitude, we can calculate the magnitude of one vector and use it for the other.
a = √(21^2 + 21^2) = 30
Therefore, the magnitude of both vectors a and b is 30.
Step 2: Calculate the angle between the two vectors.
The angle between two vectors a and b is given by the formula θ = cos^1 (a.b/ab).
Using the given values, we can calculate the angle between the two vectors as follows:
θ = cos^1 (21/30*30) = 60°
Therefore, the angle between the two vectors is 60°.
Question:
Show that each of the given three vectors is a unit vector: 71(2i^+3j^+6k^),71(3i^−6j^+2k^),71(6i^+2j^−3k^) . Also, show that they are mutually perpendicular to each other.
Answer:
 71(2i^+3j^+6k^): To show that this vector is a unit vector, we must calculate the magnitude of the vector. We can do this by using the Pythagorean Theorem:
71(2i^+3j^+6k^)=sqrt[(2^2)+(3^2)+(6^2)]=sqrt[4+9+36]=sqrt[49]=7
Therefore, the magnitude of the vector is 7, and to make it a unit vector, we must divide it by 7:
71(2i^+3j^+6k^)=7/7(2i^+3j^+6k^)=1(2i^+3j^+6k^)
 71(3i^−6j^+2k^): To show that this vector is a unit vector, we must calculate the magnitude of the vector. We can do this by using the Pythagorean Theorem:
71(3i^−6j^+2k^)=sqrt[(3^2)+(−6^2)+(2^2)]=sqrt[9+36+4]=sqrt[49]=7
Therefore, the magnitude of the vector is 7, and to make it a unit vector, we must divide it by 7:
71(3i^−6j^+2k^)=7/7(3i^−6j^+2k^)=1(3i^−6j^+2k^)
 71(6i^+2j^−3k^):To show that this vector is a unit vector, we must calculate the magnitude of the vector. We can do this by using the Pythagorean Theorem:
71(6i^+2j^−3k^)=sqrt[(6^2)+(2^2)+(−3^2)]=sqrt[36+4+9]=sqrt[49]=7
Therefore, the magnitude of the vector is 7, and to make it a unit vector, we must divide it by 7:
71(6i^+2j^−3k^)=7/7(6i^+2j^−3k^)=1(6i^+2j^−3k^)
To show that they are mutually perpendicular to each other, we must calculate the dot product of each vector. If the dot product of two vectors is 0, then they are perpendicular to each other.
71(2i^+3j^+6k^)•71(3i^−6j^+2k^)=71^2(2•3 + 3•−6 + 6•2)=71^2(6−18+12)=71^2(0)=0
71(2i^+3j^+6k^)•71(6i^+2j^−3k^)=71^2(2•6 + 3•2 + 6•−3)=71^2(12+6−18)=71^2(0)=0
71(3i^−6j^+2k^)•71(6i^+2j^−3k^)=71^2(3•6 + −6•2 + 2•−3)=71^2(18−12−6)=71^2(0)=0
Therefore, the three vectors are mutually perpendicular to each other.
Question:
Write two different vectors having same direction
Answer:
Vector 1: <3, 4>
Vector 2: <6, 8>
Question:
Write two different vectors having same magnitude
Answer:
Vector 1: (3, 4) Vector 2: (3, 4)
Question:
Compute the magnitude of the following vectors: a =i^+j^+k^;b =2i^−7j^−3k^;c =1/√3i^+1/√3j^−1/√3k^
Answer:
a. a = √(1^2 + 1^2 + 1^2) = √3
b. b = √((2^2) + (7^2) + (3^2)) = √58
c. c = √(1/3^2 + 1/3^2 + (1/3^2)) = √(2/3)
Question:
For given vectors, a =2i^−j^+2k^ and b =−i^+j^−k^, find the unit vector in the direction of the vector a +b
Answer:
Step 1: Add the two given vectors a \n and b \n.
a \n+b \n= 2i^−j^+2k^−i^+j^−k^
Step 2: Simplify the vector expression.
a \n+b \n= 3i^−2j^+k^
Step 3: Find the magnitude of the vector a \n+b \n.
a \n+b \n = \sqrt{(3i^)^2+(−2j^)^2+(k^)^2}
a \n+b \n = \sqrt{9+4+1}
a \n+b \n = \sqrt{14}
Step 4: Calculate the unit vector in the direction of the vector a \n+b \n.
\frac{a \n+b \n}{a \n+b \n} = \frac{3i^−2j^+k^}{\sqrt{14}}
\frac{a \n+b \n}{a \n+b \n} = \frac{3i^−2j^+k^}{14^{\frac{1}{2}}}
\frac{a \n+b \n}{a \n+b \n} = \frac{3}{14^{\frac{1}{2}}}i^−\frac{2}{14^{\frac{1}{2}}}j^+\frac{1}{14^{\frac{1}{2}}}k^
Therefore, the unit vector in the direction of the vector a \n+b \n is \frac{3}{14^{\frac{1}{2}}}i^−\frac{2}{14^{\frac{1}{2}}}j^+\frac{1}{14^{\frac{1}{2}}}k^.
Question:
If a is a nonzero vector of magnitude ′a′ and λ a nonzero scalar, then λa
is unit vector if
A λ=1
B λ=−1
C a=∣λ∣
D a=1/∣λ∣
Answer:
A λ=1
Question:
Find the position vector of the mid point of the vector joining the points P(2,3,4) and Q(4,1,2).
Answer:
Step 1: Find the vector joining P and Q. Vector PQ = (42, 13, 24) = (2, 2, 2)
Step 2: Find the magnitude of the vector PQ. PQ = √(2² + (2)² + (2)²) = √8
Step 3: Find the unit vector of the vector PQ. Unit vector of PQ = (2/√8, 2/√8, 2/√8)
Step 4: Find the mid point of vector PQ. Mid point of PQ = (2 + 4)/2, (3 + 1)/2, (4 + 2)/2 = (3, 2, 3)
Step 5: Find the position vector of the mid point of vector PQ. Position vector of the mid point of PQ = (3, 2, 3)
Question:
Show that the vectors 2i^−3j^+4k^ and −4i^+6j^−8k^ are collinear
Answer:
Step 1: Find the components of the two vectors.
Vector 1: 2i^−3j^+4k^ Vector 2: −4i^+6j^−8k^
Step 2: Determine the ratio of the two vectors.
Ratio = (2i^−3j^+4k^):(−4i^+6j^−8k^)
Step 3: Simplify the ratio.
Ratio = (−2i^+3j^−4k^):(−4i^+6j^−8k^)
Step 4: The ratio is equal to 1, so the vectors are collinear.
Question:
Find ∣x ∣, if for a unit vector a ,(x −a )⋅(x +a )=12
Answer required
Answer:
Step 1: We know that for a unit vector a, a⋅a = 1.
Step 2: We can rewrite the given equation as (x \n−a \n)⋅(x \n+a \n)=12 \n
Step 3: Expanding the equation, we get x \n2 + a \n2 2x \n a \n = 12 \n
Step 4: Solving for x \n, we get x \n = a \n ± √(12 \n + 2a \n2)
Step 5: Taking absolute value, we get x \n = a \n ± √(12 \n + 2a \n2)
Question:
If a ,b ,c are unit vectors such that a +b +c =0 , find the value of a ⋅b +b ⋅c +c ⋅a
Answer:
Step 1: Understand the question.
The question is asking us to find the value of the dot product of three unit vectors such that their sum is equal to zero.
Step 2: Define the dot product.
The dot product is defined as the sum of the products of the corresponding components of two vectors.
Step 3: Calculate the dot product.
a⋅b + b⋅c + c⋅a = a1b1 + a2b2 + a3b3 + b1c1 + b2c2 + b3c3 + c1a1 + c2a2 + c3a3
Step 4: Substitute in the values.
Since the three vectors are unit vectors, their components are equal to 1. Therefore, a⋅b + b⋅c + c⋅a = 1 + 1 + 1 = 3.
Question:
Find ∣b ∣, if (a +b )⋅(a −b )=8 and ∣a ∣=8∣b ∣
Answer:
Step 1: Since ∣a \n∣=8∣b \n∣, we can divide both sides by 8 to get ∣b \n∣=∣a \n∣/8.
Step 2: Substitute the value of ∣a \n∣/8 into the equation (a \n+b \n)⋅(a \n−b \n)=8 and solve for ∣b \n∣.
Step 3: We get ∣b \n∣=2.
Question:
Find the angle between the vectors i^−2j^+3k^ and 3i^−2j^+k^.
Answer:
Step 1: Find the magnitude of the two vectors.
The magnitude of the vector i^−2j^+3k^ is: √(1²+(2)²+3²) = √14
The magnitude of the vector 3i^−2j^+k^ is: √(3²+(2)²+1²) = √14
Step 2: Find the dot product of the two vectors.
The dot product of the two vectors is: (13) + (22) + (3*1) = 7
Step 3: Calculate the angle between the two vectors using the formula:
cos θ = (dot product)/(magnitude of vector 1 * magnitude of vector 2)
cos θ = 7/(√14 * √14) = 7/14
Step 4: Calculate the angle in degrees using the inverse cosine function.
θ = cos^1(7/14) = 57.3°
Question:
Find the area of the parallelogram whose adjacent sides are determined by the vector a=i^−j^+3k^ and b=2i^−7j^+k^ .
Answer:
Step 1: Find the magnitude of vectors a and b.
a = a = sqrt((1)^2 + (1)^2 + (3)^2) = sqrt(11) b = b = sqrt((2)^2 + (7)^2 + (1)^2) = sqrt(50)
Step 2: Find the dot product of vectors a and b.
a . b = (12) + (17) + (3*1) = 11
Step 3: Find the angle between vectors a and b.
angle = arccos (11/sqrt(11)*sqrt(50)) = arccos(11/50) = 0.7536
Step 4: Find the area of the parallelogram.
Area = a * b * sin(angle) = sqrt(11) * sqrt(50) * sin(0.7536) = 25.8
Question:
Find the projection of the vector i^−j^ on the vector i^+j^ .
Answer:
Answer:
 Determine the unit vector of i^+j^:
i^+j^ = (1/√2) i^ + (1/√2) j^
 Calculate the dot product of i^−j^ and (1/√2) i^ + (1/√2) j^:
i^−j^ · (1/√2) i^ + (1/√2) j^ = (1/√2) i^−j^ · i^ + (1/√2) i^−j^ · j^
 Simplify the dot product:
(1/√2) i^−j^ · i^ + (1/√2) i^−j^ · j^ = (1/√2) (1  0) + (1/√2) (0  1)
 Calculate the projection of i^−j^ on i^+j^:
Projection of i^−j^ on i^+j^ = (1/√2)  (1/√2) = 0
Question:
Find a unit vector perpendicular to each of the vector a +b and a −b , where a =3i^+2j^+2k^ and b =i^+2j^−2k^ .
Answer:

To find a unit vector perpendicular to a vector a+b, we need to find the cross product of the two vectors, a and b.

The cross product of two vectors a and b is given by a x b = (a2b3  a3b2, a3b1  a1b3, a1b2  a2b1).

Substituting the values of a and b in the above equation, we get a x b = (2(2)  (2)(2), (2)(3)  (3)(2), (3)(2)  (2)(2)) = (4, 6, 6).

To find the unit vector perpendicular to a+b, we need to divide the above vector by its magnitude. The magnitude of the vector a x b is given by a x b = sqrt(42 + (6)2 + 62) = sqrt(40).

Hence, the unit vector perpendicular to a+b is (4/sqrt(40), 6/sqrt(40), 6/sqrt(40)).

Similarly, to find a unit vector perpendicular to ab, we need to find the cross product of the two vectors, a and b.

The cross product of two vectors a and b is given by a x (b) = (a2b3  a3b2, a3b1  a1b3, a1b2  a2b1).

Substituting the values of a and b in the above equation, we get a x (b) = (2(2)  (2)(2), (2)(3)  (3)(2), (3)(2)  (2)(2)) = (4, 6, 6).

To find the unit vector perpendicular to ab, we need to divide the above vector by its magnitude. The magnitude of the vector a x (b) is given by a x (b) = sqrt(42 + 62 + (6)2) = sqrt(40).

Hence, the unit vector perpendicular to ab is (4/sqrt(40), 6/sqrt(40), 6/sqrt(40)).
Question:
Why were they named after the months of the year?
Answer:
 Who is “they” referring to?
 What is the connection between “they” and the months of the year?