Question

Given that $\overline{a},\overline{b},\overline{c}$ is a triad of three non-coplanar vectors such that,

$(1-h-k+l)\overline{a}+(2h-3k-l)\overline{b}+(3+h+k)\overline{c}=0$. Find $h,k,l$ in this case.

• A. $h=-13/5,k=-2/5,l=-4$
• B. $h=-13/5,k=-3/5,l=-2$
• C. $h=-12/5,k=-2/5,l=-4$
• D. $h=-12/5,k=-3/5,l=-2$

Answer 1

The correct option is A $h=-13/5,k=-2/5,l=-4$

Given $\overline{a},\overline{b},\overline{c}$ is a triad of three non-coplanar vectors.
$(1-h-k+l)\overline{a}+(2h-3k-l)\overline{b}+(3+h+k)\overline{c}=0$.

comparing on both sides gives
$h+k-l-1=0$ -----(1)
$2h-3k-l=0$ ----(2)
$h+k+3=0$ ----(3)

from (1) and (3)

$l=-4$

solving (2) and (3) for $h,k$ gives $h=\frac{-13}{5}$ and $k=\frac{-2}{5}$

$\therefore h=\frac{-13}{5},k=\frac{-2}{5},l=-4$

Hence, option A.