SATHEE: Vectors 4 Question 8

Vectors 4 Question 8

8. Let $\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = | \vec{c} |, | \vec{c} - \vec{a} | = 2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $30^{\circ}$ , then $∣ \vec{a} \times \vec{b}) \times \vec{c} ∣$ is equal to

(1999, 2M)

(a) $\frac{2}{3}$

(b) $\frac{3}{2}$

(d) 3

Show Answer

Answer:

Correct Answer: 8. (b)

Solution:

NOTE In this question, vector $\vec{c}$ is not given, therefore, we cannot apply the formulae of $\vec{a} \times \vec{b} \times \vec{c}$ (vector triple product)

Now, $| (\vec{a} \times \vec{b}) \times \vec{c} | = | \vec{a} \times \vec{b} | | \vec{c} | \sin 30^{\circ}$

Again, $| \vec{a} \times \vec{b} | = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 1 & 0 \end{array} | = 2 \hat{i} - 2 \hat{j} + \hat{k}$

$\Rightarrow | \vec{a} \times \vec{b} | = \sqrt{2^{2} + (- 2)^{2} + 1} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

Since,

$| \vec{c} - \vec{a} | = 2 \sqrt{2}$

[given]

$\begin{array}{lc} \Rightarrow & | \vec{c} - \vec{a} |^{2} = 8 \\ \Rightarrow & (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8 \\ \Rightarrow & \vec{c} \cdot \vec{c} - \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{a} = 8 \\ \Rightarrow & | \vec{c} |^{2} + | \vec{a} |^{2} - 2 \vec{a} \cdot \vec{c} = 8 \\ \Rightarrow & | \vec{c} |^{2} + 9 - 2 | \vec{c} | = 8 \\ \Rightarrow & | \vec{c} |^{2} - 2 | \vec{c} | + 1 = 0 \\ \Rightarrow & (| \vec{c} | - 1)^{2} = 0 \Rightarrow | \vec{c} | = 1 \end{array}$

From Eq. (i), $| (\vec{a} \times \vec{b}) \times \vec{c} | =$ (3) (1) $\cdot (\frac{1}{2}) = \frac{3}{2}$

Vectors 4 Question 8

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Vectors 4 Question 8

8. Let a→=2i^+j^−2k^ and b→=i^+j^. If c→ is a vector such that a→⋅c→=|c→|,|c→−a→|=22 and the angle between (a→×b→) and c→ is 30∘, then ∣a→×b→)×c→∣ is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu