Question

Angle between two planes $a_{1}x+b_{1}x+c_{1}x+d_1=0$ & $a_{2}x+b_{2}x+c_{2}x+d_2=0$ is given by-

• A. $tan\left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
• B. $sin\left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
• C. $cos\left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$
• D. $cot\left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$

Answer 1

The correct option is C $cos\left(\theta \right)=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$

We know that the angle between two planes is the angle between its normals.

Let the equation of one plane be r. $n_1=k_1$ and the another plane be r. $n_2=k_2$

So the angle between these normals will be -

$cos=n_1.n_2{n}_1.n_2$

In the equation of a plane in cartesian form we know that the coefficients of x,y & z give us the direction ratios of the normal vector.

So, for the first plane

$n_1=a_{1}i+b_{1}j+c_{1}k$

Similarly for the second plane the normal vector $n_2=a_{2}i+b_{2}j+c_{2}k$

$cos\theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$