Question

The vector equation of a plane, which is at a distance of \[8\]unit from the origin and which is normal to the vector\[2i + j + 2k\], is
A) \[\;r.(2i + j + k) = 24\]
B) \[r.(2i + j + 2k) = 24\]
C) \[\;r.(i + j + k) = 24\]
D) None of these

Answer 1

Hint: in this question, we have to find the vector equation of a plane which is at a distance of \[8\] unit from origin and perpendicular to given vector. It can be found by simply applying formula of plane in origin from distance and unit vector form.



Formula Used:Where
\[\overrightarrow r \]Is a position vector of any arbitrary point.
 is a unit vector
d is constant



Complete step by step solution:In question normal vector is given which is equal to \[2i + j + 2k\]and \[d = 8\]



Put this value in equation
Where
\[\overrightarrow r \]Is a position vector of any arbitrary point.
 is a unit vector
d is constant
Required equation is
\[\;r.(\dfrac{2}{3}i + \dfrac{1}{3}j + \dfrac{2}{3}k) = 8\]



Option ‘C’ is correct



Note: Here we need to remember that unit vector is ratio of given vector to modulus of given vector.
Modulus of any vector is equal to \[\sqrt {{x^2} + {y^2} + {z^2}} \].