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Vector Algebra Part 2

Section: Mathematics

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Question : 60 of 100

Marks: +1, -0

Let

\overset{\rightarrow}{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} a_i>0, i=1,2,3

be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of

\overset{\rightarrow}{a}

3 \hat{i}+4 \hat{j}

\overset{\rightarrow}{b}

be a vector obtained by rotating

\overset{\rightarrow}{a}

90^{\circ}

\overset{\rightarrow}{a}, \overset{\rightarrow}{b}

x

-axis are coplanar, then projection of a vector

\overset{\rightarrow}{b}

3 \hat{i}+4 \hat{j}

[25-Jun-2022-Shift-1]

$\sqrt{7}$
$\sqrt{2}$
2
7

Solution: 👈: Video Solution

\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma=1 \Rightarrow \cos^2 \alpha=\;\frac{1}{3} \Rightarrow \cos \alpha=\;\frac{1}{\sqrt{3}}

\overset{\rightarrow}{a}=\;\frac{\lambda}{3}(\hat{i}+\hat{j}+\hat{k}), \lambda>0

\;\frac{\lambda}{\sqrt{3}} \;\frac{(\hat{i}+\hat{j}+\hat{k})\cdot(3\hat{i}+4\hat{j})}{\sqrt{3^2+4^2}}=7

\Rightarrow \;\frac{\lambda}{\sqrt{3}}(3+4)=7 \times 5

\therefore \lambda=5 \sqrt{3}

\overset{\rightarrow}{a}=5(\hat{i}+\hat{j}+\hat{k})

Let

\overset{\rightarrow}{b}=p \hat{i}+q \hat{j}+r \hat{k}

\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{b}=0

and

\left[\overset{\rightarrow}{a} \hat{i}\right]=0

\Rightarrow p+q+r=0

\& \begin{vmatrix} p & q & r \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix}=0 \Rightarrow q=r

\overset{\rightarrow}{b}=-2 r \hat{i}+r \hat{j}+r \hat{k}

\overset{\rightarrow}{b}=r(-2 \hat{i}+\hat{j}+\hat{k})

\;\text{ Now }\;\left|\overset{\rightarrow}{a}\right|=\left|\overset{\rightarrow}{b}\right|

5 \sqrt{3}=|r| \sqrt{b} \Rightarrow |r|=\;\frac{5}{\sqrt{2}}

\Rightarrow \;\text{ Projection of }\; \overset{\rightarrow}{b}\;\text{ on }\; 3\hat{i}+4\hat{j}=\left|\frac{\overset{\rightarrow}{b}\cdot(3\hat{i}+4\hat{j})}{\sqrt{3^2+4^2}}\right|

=|r| \;\frac{(-6+4)}{5}=\left|\;\frac{-2r}{5}\right|

\;\text{ Projection }\;=\;\frac{2}{5} \times \;\frac{5}{\sqrt{2}}=\sqrt{2}

\therefore B

is correct.

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