Test Index

Vector Algebra Part 2

Section: Mathematics

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Question : 97 of 100

Marks: +1, -0

Let

\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}

\vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k}

be two vectors, such that

\vec{a} \times \vec{b} = -\hat{i} + 9 \hat{i} + 12 \hat{k}

. Then the projection of

\vec{b} - 2 \vec{a}

\vec{b} + \vec{a}

[27-Jul-2022-Shift-1]

2
$\frac{39}{5}$
9
$\frac{46}{5}$

Solution:

\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}, \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k}

\vec{a} \times \vec{b} = -\hat{i} + 9 \hat{j} + 12 \hat{k}

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4 \end{vmatrix} = -\hat{i} + 9 \hat{j} + 12 \hat{k}

4 + 5\beta = -1 \Rightarrow \beta = -1

-5\alpha - 3 = 12 \Rightarrow \alpha = -3

\vec{b} - 2\vec{a} = 3\hat{i} - 5\hat{j} + 4\hat{k} - 2(-3\hat{i} + \hat{j} - \hat{k})

\vec{b} - 2\vec{a} = 9\hat{i} - 7\hat{j} + 6\hat{k}

\vec{b} + \vec{a} = (3\hat{i} - 5\hat{j} + 4\hat{k}) + (-3\hat{i} + \hat{j} - \hat{k})

\vec{b} + \vec{a} = -4\hat{j} + 3\hat{k}

Projection of

\vec{b} - 2\vec{a}

on

\vec{b} + \vec{a}

is

= \frac{(\vec{b} - 2\vec{a}) \cdot (\vec{b} + \vec{a})}{|\vec{b} + \vec{a}|}

= \frac{28+18}{5} = \frac{46}{5}

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