Vector Algebra Question 4
Question 4 - 25 January - Shift 1
The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is
(1) $3 \sqrt{2}$
(2) 1
(3) $\sqrt{6}$
(4) $2 \sqrt{3}$
Show Answer
Answer: (1)
Solution:
Formula: Scalar product of two vectors : projection of a vector on the another vector.
$\overrightarrow{{}b}=\lambda \overrightarrow{{}a} \times(\overrightarrow{{}a} \times \hat{j})$
$\Rightarrow \overrightarrow{{}b}=\lambda(-2 \hat{i}-2 \hat{j}+2 \hat{k})$
$|\overrightarrow{{}b}|=|\overrightarrow{{}a}|$
$ \therefore \sqrt{6}=\sqrt{12}|\lambda| \Rightarrow \lambda= \pm \frac{1}{\sqrt{2}}$
$(\lambda=\frac{1}{\sqrt{2}}.$ rejected $\because \vec{b}$ makes acute angle with $y$ axis $)$
$\overrightarrow{{}b}=-\sqrt{2}(-\hat{i}-\hat{j}+\hat{k})$
$\frac{(3 \vec{a}+\sqrt{2} \vec{b}) \cdot \vec{c}}{|\vec{c}|}=3 \sqrt{2}$
