Question

if a,b,c are non coplanar find the pt of intersection of line passing thrgh 2a+3b-c; 3a+4b-2c; with line joining pts a-2b+3c; a-6b+6c

Answer 1

Please see the attachment

Answer 2

One of the Vectors Should be having Linear Combination

Step-by-step explanation:

Let the given four points be P, Q, R and S respectively.  

• If Vector are coplanar then points are coplanar.

• If one of the vectors can express itself as linear combination off the two, then the vectors are called coplanar.

So, let  

$\overrightarrow{PQ} =x\overrightarrow{PR} + y\overrightarrow{PS}$

⇒$-\overrightarrow{a} -5\overrightarrow{b}+4\overrightarrow{c}$

= $x(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c})+y(-\overrightarrow{a}-9\overrightarrow{b}-7\overrightarrow{c})$

⇒ $-\overrightarrow{a} -5\overrightarrow{b}+4\overrightarrow{c}$

= $(x-y)\overrightarrow{a}+(x-9y)\overrightarrow{b}+(-x+7y)\overrightarrow{c}$

=$x-y=-1, x-9y=-5, -x+7y = 4$

• [Equating coefficients of $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ on both sides]

Solving the first of these three equations,

we get x = $-\frac {1}{2}, y = \frac {1}{2}$

These values also satisfy the third equation which says that 4 vectors are Coplanar.