Question

The position vectors of two points P and Q are \[3i + j + 2k\]and \[i - 2j - 4k\;\] respectively. The equation of the plane through Q and perpendicular to PQ is
A) \[r.(2i + 3j + 6k) = 28\]
B) \[\;\;r.(2i + 3j + 6k) = 32\]
C) \[\;r.(2i + 3j + 6k) + 28 = 0\]
D) None of these

Answer 1

Hint: in this question, we have to find the equation of a plane passing through a given point and perpendicular to a given line. First, find the equation of line which is equal to the difference of position vectors of given points. Then use the standard equation of plane in order to find the equation of required 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 .
Formula for position vector is given by
\[\overrightarrow {AB} = Position{\rm{ }}vector{\rm{ }}of{\rm{ }}B{\rm{ }}-{\rm{ }}position{\rm{ }}vector{\rm{ }}of{\rm{ }}A\]



Complete step by step solution:Equation of line which is perpendicular to required plane is given by
\[\overrightarrow {AB} = Position{\rm{ }}vector{\rm{ }}of{\rm{ }}B{\rm{ }}-{\rm{ }}position{\rm{ }}vector{\rm{ }}of{\rm{ }}A\]
\[\overrightarrow P = 3i + j + 2k\]
\[\overrightarrow Q = i - 2j - 4k\;\]
\[\overrightarrow {PQ} = (i - 2j - 4k\;) - (3i + j + 2k)\]
\[\overrightarrow {PQ} = - 2i - 3j - 6k\]
This is an equation of line which is perpendicular to required plane
\[\overrightarrow n = \overrightarrow {PQ} = - 2i - 3j - 6k\]
Plane is passing through point \[i - 2j - 4k\;\]
\[\overrightarrow a = i - 2j - 4k\;\]
Now 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 .
Now putting value of a and n in equation \[(\overrightarrow r - \overrightarrow a ).\overrightarrow n = 0\]
We get
\[(\overrightarrow r - (i - 2j - 4k)\;).( - 2i - 3j - 6k) = 0\]
\[(\overrightarrow r \;).( - 2i - 3j - 6k) - ( - (i - 2j - 4k))( - 2i - 3j - 6k) = 0\]
On rearranging we get
\[ - \overrightarrow r .(2i + 3j + 6k) - (i - 2j - 4k).(2i + 3j + 6k) = 0\]
\[\overrightarrow r .(2i + 3j + 6k) - (2 - 6 - 24) = 0\]
\[\overrightarrow r .(2i + 3j + 6k) - ( - 28) = 0\]
Now the equation of require plane is
\[\overrightarrow r .(2i + 3j + 6k) + 28 = 0\]



Option ‘C’ is correct



Note: Here we need to remember that; vector PQ is an equation of line which is perpendicular to the required plane. Position vector is a vector which give position of a point with respect to an origin.