Vectors - Problems - Meaningful vector expressions

  • In this lecture, we will discuss problems related to vectors
  • We will learn how to express vector quantities meaningfully
  • Understanding these concepts is crucial for solving vector problems effectively

Problem 1: Addition of vectors

Consider the following vectors:

  • Vector A with magnitude 4 and direction 30 degrees from x-axis
  • Vector B with magnitude 3 and direction 60 degrees from x-axis We need to find the resultant of these vectors using vector addition.

Solution to Problem 1

To find the resultant vector, we will add the x-components and y-components separately:

  • X-component of A: 4 cos 30
  • Y-component of A: 4 sin 30
  • X-component of B: 3 cos 60
  • Y-component of B: 3 sin 60 Now, we can add the x-components and y-components to get the resultant vector.

Problem 2: Scalar multiplication

Consider the vector A with magnitude 3 and direction 45 degrees from x-axis. We need to find the vector obtained by multiplying A by a scalar value of 2.

Solution to Problem 2

To find the vector obtained by multiplying A by a scalar value of 2, we simply multiply the magnitude of vector A by 2, while keeping the direction the same. Magnitude of new vector = 3 * 2 = 6 Direction of new vector = 45 degrees Thus, the resultant vector is 6 at an angle of 45 degrees from x-axis.

Problem 3: Dot product of vectors

Consider two vectors A and B.

  • Vector A has a magnitude of 5 and is directed 30 degrees to the x-axis.
  • Vector B has a magnitude of 4 and is directed 60 degrees to the x-axis. We need to find the dot product of these two vectors.

Solution to Problem 3

To find the dot product of two vectors, we multiply the magnitudes of the vectors and the cosine of the angle between them. Dot product = |A| * |B| * cos(theta) = 5 * 4 * cos(30 - 60) Simplifying further, we get: Dot product = 20 * cos(-30) Dot product = 20 * (sqrt(3)/2) Dot product = 10 * sqrt(3) Thus, the dot product of vectors A and B is 10 * sqrt(3).

Problem 4: Projection of vectors

Consider two vectors A and B.

  • Vector A has a magnitude of 3 and is directed 30 degrees to the x-axis.
  • Vector B has a magnitude of 4 and is directed 45 degrees to the x-axis. We need to find the projection of vector A on vector B.

Solution to Problem 4

To find the projection of vector A on vector B, we need to use the formula: Projection of A on B = |A| * cos(theta) Substituting the given values, we get: Projection of A on B = 3 * cos(30 - 45) Projection of A on B = 3 * cos(-15) Projection of A on B = 3 * (sqrt(6) + sqrt(2))/4 Thus, the projection of vector A on vector B is (3 * (sqrt(6) + sqrt(2))/4).

Problem 5: Cross product of vectors

Consider two vectors A and B.

  • Vector A has a magnitude of 2 and is directed 45 degrees to the x-axis.
  • Vector B has a magnitude of 3 and is directed 60 degrees to the x-axis. We need to find the cross product of these two vectors.
  • The cross product of two vectors results in a vector perpendicular to both vectors.
  1. Solution to Problem 5 To find the cross product of two vectors A and B, we need to use the formula: Cross product = |A| * |B| * sin(theta) * n Where n is the unit vector perpendicular to the plane of A and B.
  • Since A and B are directed along the x-y plane, the cross product will be along the z-axis.
  1. Calculating the cross product Cross product = 2 * 3 * sin(45 - 60) * k Simplifying further, we get: Cross product = 6 * sin(-15) * k
  1. Using trigonometric identity We can use the trigonometric identity sin(-x) = -sin(x) to simplify the expression further: Cross product = 6 * (-sin(15)) * k
  1. Substituting the value To calculate the cross product value, we can substitute the approximate value of sin(15) as 0.2588: Cross product = 6 * (-0.2588) * k
  1. Simplifying the expression Cross product ≈ -1.5528 * k
  1. Resultant vector Thus, the cross product of vectors A and B is approximately -1.5528 * k.
  • This vector is perpendicular to both A and B and is directed along the z-axis.
  1. Problem 6: Magnitude of a vector Consider a vector A with components (2, -5, 3).
  • We need to find the magnitude of vector A using the Pythagorean theorem.
  1. Solution to Problem 6 To find the magnitude of a vector, we use the formula: Magnitude = sqrt(x^2 + y^2 + z^2)
  1. Calculating the magnitude Magnitude = sqrt(2^2 + (-5)^2 + 3^2) Simplifying further, we get: Magnitude = sqrt(4 + 25 + 9) Magnitude = sqrt(38) Thus, the magnitude of vector A is sqrt(38).

Problem 7: Cartesian form of a vector

Consider a vector A with components (3, -2, 6).

  • We need to express vector A in its Cartesian form. Solution to Problem 7 The Cartesian form of a vector is given by the expression A = (x, y, z), where x, y, and z are the components of the vector. For vector A, the Cartesian form is: A = (3, -2, 6)
  1. Problem 8: Parametric form of a vector Consider a vector A with components (2, -1, 3).
  • We need to express vector A in its parametric form. Solution to Problem 8 The parametric form of a vector is given by the expression A = ai + bj + ck, where a, b, and c are the coefficients of i, j, and k respectively. For vector A, the parametric form is: A = 2i - j + 3k
  1. Problem 9: Vector equation of a line Consider a vector A with components (1, -2, 3) and a point P(2, 1, -3) lying on the line.
  • We need to find the vector equation of the line passing through point P and parallel to vector A. Solution to Problem 9 The vector equation of a line passing through a point P and parallel to vector A is given by the expression r = a + tb, where r is the position vector, a is the position vector of point P, t is a scalar parameter, and b is the direction vector of the line. For this problem, the vector equation of the line is: r = <2, 1, -3> + t<1, -2, 3>
  1. Problem 10: Angle between two vectors Consider two vectors A and B with components A(2, 3, -1) and B(-1, 2, 4).
  • We need to find the angle between vectors A and B. Solution to Problem 10 To find the angle between two vectors A and B, we can use the dot product formula: cos(theta) = (A dot B) / (|A| * |B|) So, let’s calculate the dot product and magnitudes: A dot B = (2 * -1) + (3 * 2) + (-1 * 4) |A| = sqrt(2^2 + 3^2 + (-1)^2) |B| = sqrt((-1)^2 + 2^2 + 4^2) Substituting the values, we can find cos(theta) and then calculate the angle.
  1. Problem 11: Unit vector Consider a vector v with components (4, -3, 5).
  • We need to find the unit vector in the direction of v. Solution to Problem 11 The unit vector in the direction of v can be found by dividing each component of the vector by its magnitude. First, calculate the magnitude of vector v: |v| = sqrt(4^2 + (-3)^2 + 5^2) Then, divide each component of v by its magnitude to obtain the unit vector. Unit vector = (4/|v|, -3/|v|, 5/|v|)
  1. Problem 12: Collinear vectors Consider two vectors A and B with components A(1, 2, -3) and B(-2, -4, 6).
  • We need to determine if vectors A and B are collinear (parallel, anti-parallel, or neither). Solution to Problem 12 To determine if two vectors A and B are collinear, we need to check if there exists a scalar k such that B = kA. We can divide the corresponding components of A and B to find the scalar ratio: -2/1 = -4/2 = 6/-3 If the ratio is constant, then vectors A and B are collinear. Since the ratio is constant (-2/1 = -4/2 = 6/-3 = -2), vectors A and B are collinear.
  1. Problem 13: Coplanar vectors Consider three vectors A, B, and C with components A(1, 2, -1), B(-2, 0, 5), and C(3, 1, -4).
  • We need to determine if vectors A, B, and C are coplanar (lie on the same plane) or not. Solution to Problem 13 To determine if three vectors A, B, and C are coplanar, we can use the scalar triple product. If the scalar triple product of A, B, and C is zero, then the vectors are coplanar. Scalar triple product = A dot (B cross C) If the scalar triple product equals zero, then the vectors are coplanar. Otherwise, they are not coplanar. Let’s calculate the scalar triple product to determine if A, B, and C are coplanar.
  1. Problem 14: Perpendicular vectors Consider two vectors A and B with components A(1, 2, -3) and B(2, -4, 2).
  • We need to determine if vectors A and B are perpendicular (orthogonal) or not. Solution to Problem 14 To determine if two vectors A and B are perpendicular, we can check if their dot product is zero. If A dot B = 0, then vectors A and B are perpendicular. Let’s calculate the dot product of vectors A and B to determine if they are perpendicular. (A dot B) = (1 * 2) + (2 * -4) + (-3 * 2) If the dot product equals zero, then vectors A and B are perpendicular. Otherwise, they are not perpendicular.
  1. Problem 15: Angle between vectors given their dot product Consider two vectors A and B with components A(3, 4) and B(5, -2).
  • We are given that the dot product of vectors A and B is 11. We need to find the angle between the vectors. Solution to Problem 15 The dot product of two vectors A and B can be used to find the angle between them by using the formula: cos(theta) = (A dot B) / (|A| * |B|) Substituting the given values, we can calculate cos(theta) and then find the angle between vectors A and B using inverse cosine (acos) function. cos(theta) = 11 / (sqrt(3^2 + 4^2) * sqrt(5^2 + (-2)^2)) theta = acos(11 / (sqrt(25) * sqrt(29))) Thus, the angle between vectors A and B is approximately 0.698 radians or 40 degrees.

Recap

  • In this lecture, we discussed various problems related to vectors.
  • We learned how to add vectors, subtract vectors, multiply vectors by scalars, calculate dot products, and find the projection of a vector on another vector.
  • We also covered topics such as expressing vectors in Cartesian and parametric forms, finding the magnitude of a vector, determining the unit vector in a given direction, and solving problems related to collinearity, coplanarity, perpendicularity, and angles between vectors.
  • Understanding these concepts will help us solve vector problems effectively in the 12th Boards exam.