The direction ratios of normal to the plane, $L_1: a_1x + b_1y + c_1z = 0,$
are $a_1, b_1, c_1$ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between $L_1$ and $L_2$ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the planes are are $7x + 5y + 6z + 30 = 0$ and
$3x - y - 10z + 4 = 0$
Here, $a_1 = 7, b_1 = 5, c_1 = 6$
$a_2 = 3, b_2 = -1, c_2 = -10$
$a_1a_2 + b_1b_2 + c_1c_2 = 7 \times 3 + 5 \times (-1) + 6 \times (-10)$
$=-44\neq0$
Therefore, the given planes are not perpendicular.
$\frac{\text{a}_1}{\text{a}_2}=\frac{7}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{5}{-1}=-5,\ \frac{\text{c}_1}{\text{c}_2}=\frac{6}{-10}=\frac{3}{-5}$
It can be seen that, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2},$
Therefore, the given planes are not parallel.
The angle between them is given by,
$\text{Q}=\cos^{-1}\begin{vmatrix}\frac{7\times3+5\times(-1)+6\times(-10)}{\sqrt{(7)^2+(5)^2+(6)^2\times\sqrt{(3)^2+(-1)^2+(-10)^2}}}\end{vmatrix}$
$=\cos^{-1}\Bigg|\frac{21-5-60}{\sqrt{110}\times\sqrt{110}}\Bigg|$
$=\cos^{-1}\frac{44}{110}$
$=\cos^{-1}=\frac{2}{5}.$

Three Dimensional Geometry — MATHS STD 12 Science — Question

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