Which one of the following planes is normal to the plane $$3x + y + z = 5\,?$$
A.
$$x + 2y + z = 6$$
B.
$$x - 2y + z = 6$$
C.
$$x + 2y - z = 6$$
D.
$$x - 2y - z = 6$$
Answer :
$$x - 2y - z = 6$$
Solution :
Direction cosines of the normal to the plane $$3x + y + z = 5$$ are $$3,\,1,\,1$$
Direction cosines of the normal to the plane $$x - 2y - z = 6$$ are $$1,\,- 2,\, - 1$$
Sum of the product of direction cosines $$ = 3 \times 1 + 1 \times \left( { - 2} \right) + 1 \times \left( { - 1} \right) = 0$$
Hence, normls to the two planes are perpendicular to each other. Therefore two planes are also perpendicular.
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :