Question

A perpendicular is drawn from a point lying on $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{1}$ to the plane x + y + z = 3 such that the foot of the perpendicular Q lies on the plane x – y + z = 3. Then the coordinate of Q are
( a ) ( 2, 0, 1 )
( b ) ( 4, 0, -1 )
( c ) ( -1, 0, 4 )
( d ) ( 1, 0, 2 )

Answer 1

Hint: Firstly we will find the general point on line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{1}=k$ in terms of k and then using concept of perpendicular line and plane condition that is \[\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}=-\dfrac{(a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\], we will find coordinates of Q in terms of k and then we will find value of k by putting coordinates of Q in equation x – y + z = 3, then we will substitute values of k in coordinates of Q in terms of k to get numerical coordinates of Q

Complete step-by-step answer:
We know that, the standard form of equation $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line.
Now, we are given that perpendicular is drawn from a point lying on $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{1}$ to the plane x + y + z = 3 such that the foot of the perpendicular Q lies on the plane x – y + z = 3.
So, let point P be the point on line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{1}=k$, then coordinates of the point P of form P( 2k + 1, -k – 1, k ) where, we have point x =2k + 1, y = -k – 1, z = k .
Now, as PQ is the perpendicular where Q lies on the plane x + y +z = 3
So, let coordinates of Q be $Q\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$
Now, if line AB and plane ax + by + cz + d = 0 are perpendicular then, we have
\[\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}=-\dfrac{(a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\], where (\[{{x}_{1}}\],\[{{y}_{1}}\],\[{{z}_{1}}\]) is coordinate of point B on both plane and straight line and (\[{{x}_{2}}\],\[{{y}_{2}}\],\[{{z}_{2}}\]) is coordinate of point A.
Now, as line PQ and equation of plane are perpendicular to each other so, we can say that
\[\dfrac{{{x}_{1}}-2k-1}{1}=\dfrac{{{y}_{1}}+k+1}{1}=\dfrac{{{z}_{1}}+k}{1}=-\dfrac{(2k\text{ }+\text{ }1+(-k\text{ }1)+k)}{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}\], where \[{{x}_{1}}\],\[{{y}_{1}}\],\[{{z}_{1}}\] are coordinates of Q and \[{{x}_{2}}+{{y}_{2}}+{{z}_{2}}-3\]is equation plane to which line PQ are perpendicular.
On solving we get,
\[\dfrac{{{x}_{1}}-2k-1}{1}=\dfrac{{{y}_{1}}+k+1}{1}=\dfrac{{{z}_{1}}+k}{1}=\dfrac{3-2k}{3}\]
Now, comparing x coordinate of Q, we get
\[\dfrac{{{x}_{1}}-2k-1}{1}=\dfrac{3-2k}{3}\]
On solving we get,
\[{{x}_{1}}=\dfrac{6+4k}{3}\]
Now, comparing y coordinate of Q, we get
\[\dfrac{{{y}_{1}}+k+1}{1}=\dfrac{3-2k}{3}\]
On solving we get,
\[{{y}_{1}}=-\dfrac{5k}{3}\]
Now, comparing y coordinate of Q, we get
\[\dfrac{{{z}_{1}}+k}{1}=\dfrac{3-2k}{3}\]
On solving we get,
\[{{z}_{1}}=\dfrac{3+k}{3}\]
So, we have $Q\left( \dfrac{6+4k}{3},-\dfrac{5k}{3},\dfrac{3+k}{3} \right)$ as \[{{x}_{1}}\],\[{{y}_{1}}\],\[{{z}_{1}}\] are coordinates of Q.
Now, as point Q lies on plane x - y + z = 3, so point Q will satisfy the equation plane, then
$\dfrac{6+4k}{3}-\left( -\dfrac{5k}{3} \right)+\dfrac{3+k}{3}=3$
On solving we get
6 + 4k + 5k + 3 + k = 9
Or, k = 0
So, coordinates of Q will be $Q\left( \dfrac{6+4(0)}{3},-\dfrac{5(0)}{3},\dfrac{3+(0)}{3} \right)$
Or, Q(2, 0, 1 )

So, the correct answer is “Option A”.

Note: Always convert the equation of line firstly to standard line equation which is $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line. Also, remember that if line AB and plane ax + by + cz + d = 0 are perpendicular then, we have
\[\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}=-\dfrac{(a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\], where (\[{{x}_{1}}\],\[{{y}_{1}}\],\[{{z}_{1}}\]) is coordinate of point B on both plane and straight line and (\[{{x}_{2}}\],\[{{y}_{2}}\],\[{{z}_{2}}\]) is coordinate of point A. Avoid calculation error while solving the question.