Question

A vector n of magnitude 8 units is inclined to x-axis at ${45^0}$, y-axis at ${60^0}$ and an acute angle with z-axis. If a plane passes through a point $(\sqrt 2 , - 1,1)$ and is normal to, then its equation in vector form is
A) $r.(\sqrt 2 i + j + k) = 4$
B) $r.(\sqrt 2 i + j + k) = 2$
C) $r.(i + j + k) = 4$
D) None of these

Answer 1

Hint: in this question we have to find vector equation of plane which is inclined to axes and passing through given point. Direction of cosines of normal vector is used to find the angle which plane makes with z-axis. By using direction cosine find vector normal to the plane.



Formula Used:Equation of required plane is given by
$(\overrightarrow r - \overrightarrow a ).\overrightarrow n = 0
Where
\overrightarrow r Is a position vector of any arbitrary point.
\overrightarrow n normal vector to the plane$ .



Complete step by step solution:Suppose $\gamma$ is a angle which vector n made with z-axis
Direction of cosine of vector n is given by
$l = \cos (45)$
$m = \cos (60)$
$n = \cos (\gamma )$
We know that
${l^2} + {m^2} + {n^2} = 1$
$\dfrac{1}{2} + \dfrac{1}{4} + {n^2} = 1$
${n^2} = \dfrac{1}{4}$
$n = \dfrac{1}{2} $ Because $\gamma$ is an acute angle
In question modulus of vector n is given as 8
$n = |n|(li + mj + nk)$
$n = 8(\dfrac{1}{{\sqrt 2 }}i + \dfrac{1}{2}j + \dfrac{1}{2}k)$
$n = 4\sqrt 2 i + 4j + 4k$
We know that required plane is passing through$(\sqrt 2 , - 1,1)$
$a = \sqrt 2 i - j + k$
Equation of a plane is given by
$(\overrightarrow r - \overrightarrow a ).\overrightarrow n = 0$
Where
$\overrightarrow r Is a position vector of any arbitrary point.$
$\overrightarrow n normal vector to the plane .$
Put value of a and n in equation$(\overrightarrow r - \overrightarrow a ).\overrightarrow n = 0$
r.n = a.n
$r.(4\sqrt 2 i + 4j + 4k) = (\sqrt 2 i - j + k).(4\sqrt 2 i + 4j + 4k)$
On simplification we get equation of plane
$r.(\sqrt 2 i + j + k) = 2$;



Option ‘B’ is correct



Note: Here we should remember that, we can find the equation of any vector if its magnitude and angle which it makes with axes is given.
We should remember that sum of square of direction cosine is equal to one