Question

Let $a = i +2 j + k , b = i - j + k , c = i + j - k$ A vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is

• A. 4 i + j -4 k
• B. 3 i + j -3 k
• C. 4 i - j +4 k
• D. 2 i + j +2 k

Answer 1

Answer: (D)

Since, vectors $a$ and $b$ are in a same plane.
$ \therefore r = a +t b $
$=( i +2 j + k )+t( i - j + k ) $
$4=(1+t) i +(2-t) j +(1+t) k $ .....(i)
$\therefore $ Projection of $r$ on $c =\frac{ r \cdot c }{| c |}$
$\frac{1}{\sqrt3}=\frac{[\{(1+t) i +(2-t) j +(1+t) k \} .( i + j - k )]}{\sqrt{1^{2}+1^{2}+(-1)^{2}}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1+t+2-t-(1+t)}{\sqrt{3}}$
$\Rightarrow 1=2-t$
$\Rightarrow t=1$
On putting $t=1$ in Eq. (i), we get
$r =2 i + j +2 k$