Question

Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$, $a_i>0,\ i=1, 2, 3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\vec a$ on the vector $3\hat i+4\hat j$ be 7. Let $\vec b$ be a vector obtained by rotating $\vec a$ with 90°. If $\vec a,\vec b$ and x-axis are co-planar, then projection of a vector $\vec b$ on $3\hat i+4\hat j$ is equal to :

• A. $\sqrt 7$
• B. $\sqrt 2$
• C. $2$
• D. $7$

Answer 1

Answer: (B)

$\sqrt 2$

Answer 2

$cos^2α+cos^2β+cos^2⁡γ=1$

⇒ $cos^2⁡α=\frac 13$

$⇒cos\ ⁡α=\frac {1}{\sqrt 3}$

$\vec a=\frac {λ}{3}(\hat i+\hat j+\hat k),\ λ>0$

$\frac {λ}{\sqrt 3}\frac {(\hat i+\hat j+\hat k)⋅(3\hat i+4\hat j)}{\sqrt {3^2+4^2}}=7$

$⇒\frac {λ}{\sqrt 3}(3+4)=7×5$

∴ $λ=5\sqrt 3$

$\vec a=5(\hat i+\hat j+\hat k)$
Let
$\vec b=p\hat i+q\hat j+r\hat k$
$\vec a⋅\vec b=0$ and $[\vec a \vec b \hat i]=0$
$⇒ p + q + r = 0$ …(i)
And
$\begin{vmatrix} p & q & r \\\ 1 & 1 & 1 \\\ 1 & 0 & 0 \end{vmatrix}=0$
$⇒\vec b=−2r\hat i+r\hat j+r\hat k$
$⇒\vec b=r(−2\hat i+\hat j+\hat k)$
Now
$|\vec a|=|\vec b|$
$5\sqrt 3=|r|\sqrt b$
$|r|=5\sqrt 2$
Projection of $\vec b$ on $3\hat i+4\hat j$
= $|\frac {\vec b⋅(3\hat i+4\hat j)}{\sqrt {3^2+4^2}}|$

=$|r|\frac {(−6+4)}{5}$

=$|−\frac 25|$
Projection =$\frac 25×\frac {5}{\sqrt 2}$
Projection = $\sqrt 2$

So, the correct option is (b): $\sqrt 2$