Question

Let $\vec {a} = \hat i + \hat j + \hat k $ , $ \vec {b} = \hat {i} - \hat {j} + \hat{k} $ and $\vec {c} = \hat {i} - \hat {j} - \hat{k} $ be three vectors. $ A $ vector $ v $ in the plane of $ a $ and $ b $ , whose projection on $ c $ is $ \frac{1}{\sqrt{3}} $ is given by

• A. $ 3 \hat {i} + \hat{j} - 3\hat{k} $
• B. $ 3\hat {i} - \hat{j} + 3\hat{k} $
• C. $ 3 \hat {i} + \hat {j} + 3\hat {k} $
• D. $ -3 \hat {i} + \hat {j} + 3 \hat {k} $

Answer 1

Answer: (B)

Since, $v$ is the coplanar to $a$ and $b$
$\therefore v=a+t b^{-1}$
$=(i+j+k)+t(i-j+k)$
$\Rightarrow r = (1 + t)i + (1 - t) j + (1 + t)k .....$ (i)
$\Rightarrow r=(1+t) l+(1-t) j+(1+t) k$ (given)
$\Rightarrow \frac{ v \cdot c }{|c|}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{|(1+t) 1-1(1-t)-1(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow 1+t-1+t-1-t=1$
$\Rightarrow t=2$
On putting the value of $t$ in Eq. (i), we get
$ r =3i+(-1)j+(3) k$
$\Rightarrow v=3i-j+3 k$