If a, b and c are non-coplanar vectors and the four points with p… - Vidyadip

MCQ

If $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors and the four points with position vectors $2 \bar{a}+3 \bar{b}-\bar{c}$, $\overline{ a }-2 \overline{b}+3 \overline{ c }, \quad 3 \overline{ a }+4 \overline{b}-2 \overline{ c }$ and $k \overline{ a }-6 \overline{b}+6 \overline{ c }$ are coplanar, then $k =$

A
$0$
✓
1
C
2
D
3

✓

Answer

Correct option: B.

1

(B) Let $\bar{s}=2 \bar{a}+3 \bar{b}-\bar{c}, \bar{t}=\bar{a}-2 \bar{b}+3 \bar{c}$,
$\overline{ u }=3 \overline{ a }+4 \overline{b}-2 \overline{ c }, \overline{ v }= k \overline{ a }-6 \overline{b}+6 \overline{ c }$
$\begin{aligned} \therefore & \overline{ ST }=-\overline{ a }-5 \overline{b}+4 \overline{ c }, \overline{ SU }=\overline{ a }+\overline{ b }-\overline{ c } \\ & \overline{ SV }=( k -2) \overline{ a }-9 \overline{b}+7 \overline{ c }\end{aligned}$
Since the given points are coplanar, $[\overline{ ST }$ $\overline{ SU }$ $\overline{ SV }]=0$
$\Rightarrow\left|\begin{array}{ccc}-1 & -5 & 4 \\ 1 & 1 & -1 \\ k-2 & -9 & 7\end{array}\right|=0$
$\Rightarrow 2+5(7+ k -2)+4(-9- k +2)=0 \\ \Rightarrow 2+25+5 k -28-4 k =0 \\ \Rightarrow-1+ k =0 \\ \Rightarrow k =1$

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