Question

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