Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 10 of 72

Marks: +1, -0

Let

\vec{a}=\hat{i}-2\hat{j}+3\hat{k}, \vec{b}=2\hat{i}+\hat{j}-\hat{k}, \vec{c}=\lambda \hat{i}+\hat{j}+\hat{k}

\vec{v}=\vec{a}\times \vec{b}

\vec{v}\cdot \vec{c}=11

\vec{b}

\vec{c}

p

9 p^2

Solution:

Concept:

The vector cross product

\vec{a}\times\vec{b}

gives a vector perpendicular to both. Dot product with

\vec{c}

helps find an unknown scalar. The length of projection of

\vec{b}

\vec{c}

|\vec{b}\cdot\vec{c}|/|\vec{c}|

Explanation:

Given

\vec{a}=\hat{i}-2\hat{j}+3\hat{k}

\vec{b}=2\hat{i}+\hat{j}-\hat{k}

\vec{c}=\lambda \hat{i}+\hat{j}+\hat{k}

, and

\vec{v}=\vec{a}\times \vec{b}

Compute

\vec{v}

using determinant:

\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix}

\vec{v} = \hat{i}((-2)(-1)-3\cdot1) - \hat{j}(1\cdot(-1)-3\cdot2) + \hat{k}(1\cdot1-(-2)\cdot2)

\vec{v} = \hat{i}(2-3) - \hat{j}(-1-6) + \hat{k}(1+4)

\vec{v} = -\hat{i} + 7\hat{j} + 5\hat{k}

Given

\vec{v}\cdot\vec{c}=11

(-\hat{i}+7\hat{j}+5\hat{k})\cdot(\lambda\hat{i}+\hat{j}+\hat{k}) = 11

-\lambda + 7 + 5 = 11 \implies -\lambda + 12 = 11 \implies \lambda = 1

Thus

\vec{c} = \hat{i}+\hat{j}+\hat{k}

The projection length

p

\vec{b}

\vec{c}

p = \dfrac{|\vec{b}\cdot\vec{c}|}{|\vec{c}|}

\vec{b}\cdot\vec{c} = (2\hat{i}+\hat{j}-\hat{k})\cdot(\hat{i}+\hat{j}+\hat{k}) = 2+1-1 = 2

|\vec{c}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}

p = \dfrac{2}{\sqrt{3}}

and

p^2 = \dfrac{4}{3}

Then

9p^2 = 9 \times \dfrac{4}{3} = 12

Answer:

9p^2 = 12

and the correct option is D.

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