Vector Algebra Question 10
Question 10 - 29 January - Shift 2
Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and projection of $\overrightarrow{{}c}$ on $\overrightarrow{{}a}$ is 1 , then the projection of $\vec{c}$ on $\vec{b}$ equals :
(1) $\frac{5}{\sqrt{2}}$
(2) $\frac{1}{5}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\frac{3}{\sqrt{2}}$
Show Answer
Answer: (1)
Solution:
Formula: Scalar product of two vectors; projection of a vector on the another vector, Cross product of vectors
$ \overrightarrow{{}a} \times \overrightarrow{{}b}=15 \hat{i}-20 \hat{j}-25 \hat{k} $
Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$
$ \begin{matrix} \Rightarrow & 15 x-20 y-25 z+25=0 \\ \Rightarrow & 3 x-4 y-5 z=-5 \end{matrix} $
Also $x+y+z=4$
and $\frac{\vec{c} \cdot \vec{a}}{|\overrightarrow{{}a}|}=1 \Rightarrow 4 x+3 y=5$
$ \Rightarrow \quad \overrightarrow{{}c}=2 \hat{i}-\hat{j}+3 \hat{k} $
Projection of $\overrightarrow{{}c}$ on $\overrightarrow{{}b}=\frac{25}{5 \sqrt{2}}=\frac{5}{\sqrt{2}}$
