
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
163.8k+ views
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 \[\sqrt {{x^2} + {y^2} + {z^2}} \].
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 \[\sqrt {{x^2} + {y^2} + {z^2}} \].
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