The equation of the plane which contain the points (0, 6, 0) and (-2, -3, 4) and which is parallel to the ray with direction ratios (2, 3, -2) is:

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This question was previously asked in

AAI ATC Junior Executive 21 Feb 2023 Shift 1 Official Paper

Answer 1

Given:

The plane which contain the points (0, 6, 0) and (-2, -3, 4)

and parallel to the ray with direction ratios (2, 3, -2)

Concept:

The equation of a plane which passed through a point \(\rm (x_1,y_1,z_1)\) with normal vector is \(\rm (a,b,c)\)

\(\rm a(x-x_1)+b(y-y_1)+c(z-z_1)=0\)

Calculation:

The equation of the plane passes through (0,6,0) is

\(\rm ax+b(y-6)+cz=0......(1)\)

This plane passes through (-2,-3,4) then

\(\rm a(-2)+b(-3-6)+c(4)=0\)

\(\rm \implies-2a-9b+4c=0.......(2)\)

Plain (1) is parallel to the ray with direction ratios (2, 3, -2)

\(\rm 2a+3b-2c=0......(3)\)

Now, solving the equations (1), (2) and (3)

\(\rm \begin{vmatrix} x & y-6& z\\ -2 & -9 & 4\\ 2 & 3 & -2 \end{vmatrix} =0\)

\(\rm \implies x(18-12)-(y-6)(4-8)+z(-6+18)=0\)

\(\rm \implies 6x+4y-24+12z=0\)

\(\rm \implies 3x+2y+6z-12=0\)

Hence the option (2) is correct.