# The orthogonal projection of the point A with position vector (1,2,3) on the plane 3x-y+4z=0 is

[a] \[\left( -1,3,-1 \right)\]

[b] \[\left( -\dfrac{1}{2},\dfrac{5}{2},1 \right)\]

[c] \[\left( \dfrac{1}{2},\dfrac{-5}{2},-1 \right)\]

[d] \[\left( 6,-7,-5 \right)\]

Last updated date: 23rd Mar 2023

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Answer

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Hint: The normal vector of the plane ax+by+cz = d is (a,b,c). So choose the point P(x,y,z) PA is parallel to the normal vector. P also satisfies the plane equation. This will give you a system of three equations. Solve the system using any method. This will give the coordinates of the point P.

Complete step-by-step answer:

We know that the normal vector of the plane ax+by+cz = d is (a,b,c)

Here a = 3, b = -1, c = 4 and d = 0.

Hence the normal vector(N) of the plane is \[3\widehat{i}\text{ - }\widehat{j}\text{ + }4\widehat{k}\]

Let P(x,y,z) be the project of A on the plane 3x-y+4z=0

Since P lies on the plane, we have

3x-y+4z = 0 (i)

Also \[\overrightarrow{AP}=\left( x-1 \right)\widehat{i}\text{ +}\left( y-2 \right)\widehat{j}\text{ + }\left( z-3 \right)\widehat{k}\]

Since $AP\parallel N$we have

$\begin{align}

& \dfrac{x-1}{3}=\dfrac{y-2}{-1}=\dfrac{z-3}{4}=t\text{ (say)} \\

& \Rightarrow x=3t+1,y=2-t,z=4t+3 \\

\end{align}$

Now we have

Put the value of x,y and z in equation (i) we get

\[\begin{align}

& 3\left( 3t+1 \right)-\left( 2-t \right)+4\left( 4t+3 \right)=0 \\

& \Rightarrow 9t+3-2+t+16t+12=0 \\

& \Rightarrow 26t+13=0 \\

\end{align}\]

Subtracting 13 from both sides, we get

$\begin{align}

& 26t+13-13=0-13 \\

& \Rightarrow 26t=-13 \\

\end{align}$

Dividing both sides by 26, we get

$\begin{align}

& \dfrac{26t}{26}=\dfrac{-13}{26} \\

& \Rightarrow t=-\dfrac{1}{2} \\

\end{align}$

Hence we have

$\begin{align}

& x=3t+1=\dfrac{-3}{2}+1=\dfrac{-1}{2} \\

& y=2-t=2-\left( -\dfrac{1}{2} \right)=2+\dfrac{1}{2}=\dfrac{5}{2} \\

& z=4t+3=4\left( -\dfrac{1}{2} \right)+3=-2+3=1 \\

\end{align}$

Hence $P\equiv \left( \dfrac{-1}{2},\dfrac{5}{2},1 \right)$ is the point of the orthogonal projection of A.

Note: Alternatively we have, the equation of the line perpendicular to the plane passing through A in parametric form is $x=3t+1,y=-t+2,z=4t+3$ where t is the parameter.

The line intersects the plane at point P(t)

Then we have

\[\begin{align}

& 3\left( 3t+1 \right)-\left( 2-t \right)+4\left( 4t+3 \right)=0 \\

& \Rightarrow 9t+3-2+t+16t+12=0 \\

& \Rightarrow 26t+13=0 \\

& \Rightarrow 26t=-13 \\

& \Rightarrow t=-\dfrac{1}{2} \\

\end{align}\]

Hence

$\begin{align}

& P\equiv \left( 3\times \dfrac{-1}{2}+1,-\dfrac{-1}{2}+2,4\times \dfrac{-1}{2}+3 \right) \\

& \Rightarrow P\equiv \left( \dfrac{-1}{2},\dfrac{5}{2},1 \right) \\

\end{align}$

Complete step-by-step answer:

We know that the normal vector of the plane ax+by+cz = d is (a,b,c)

Here a = 3, b = -1, c = 4 and d = 0.

Hence the normal vector(N) of the plane is \[3\widehat{i}\text{ - }\widehat{j}\text{ + }4\widehat{k}\]

Let P(x,y,z) be the project of A on the plane 3x-y+4z=0

Since P lies on the plane, we have

3x-y+4z = 0 (i)

Also \[\overrightarrow{AP}=\left( x-1 \right)\widehat{i}\text{ +}\left( y-2 \right)\widehat{j}\text{ + }\left( z-3 \right)\widehat{k}\]

Since $AP\parallel N$we have

$\begin{align}

& \dfrac{x-1}{3}=\dfrac{y-2}{-1}=\dfrac{z-3}{4}=t\text{ (say)} \\

& \Rightarrow x=3t+1,y=2-t,z=4t+3 \\

\end{align}$

Now we have

Put the value of x,y and z in equation (i) we get

\[\begin{align}

& 3\left( 3t+1 \right)-\left( 2-t \right)+4\left( 4t+3 \right)=0 \\

& \Rightarrow 9t+3-2+t+16t+12=0 \\

& \Rightarrow 26t+13=0 \\

\end{align}\]

Subtracting 13 from both sides, we get

$\begin{align}

& 26t+13-13=0-13 \\

& \Rightarrow 26t=-13 \\

\end{align}$

Dividing both sides by 26, we get

$\begin{align}

& \dfrac{26t}{26}=\dfrac{-13}{26} \\

& \Rightarrow t=-\dfrac{1}{2} \\

\end{align}$

Hence we have

$\begin{align}

& x=3t+1=\dfrac{-3}{2}+1=\dfrac{-1}{2} \\

& y=2-t=2-\left( -\dfrac{1}{2} \right)=2+\dfrac{1}{2}=\dfrac{5}{2} \\

& z=4t+3=4\left( -\dfrac{1}{2} \right)+3=-2+3=1 \\

\end{align}$

Hence $P\equiv \left( \dfrac{-1}{2},\dfrac{5}{2},1 \right)$ is the point of the orthogonal projection of A.

Note: Alternatively we have, the equation of the line perpendicular to the plane passing through A in parametric form is $x=3t+1,y=-t+2,z=4t+3$ where t is the parameter.

The line intersects the plane at point P(t)

Then we have

\[\begin{align}

& 3\left( 3t+1 \right)-\left( 2-t \right)+4\left( 4t+3 \right)=0 \\

& \Rightarrow 9t+3-2+t+16t+12=0 \\

& \Rightarrow 26t+13=0 \\

& \Rightarrow 26t=-13 \\

& \Rightarrow t=-\dfrac{1}{2} \\

\end{align}\]

Hence

$\begin{align}

& P\equiv \left( 3\times \dfrac{-1}{2}+1,-\dfrac{-1}{2}+2,4\times \dfrac{-1}{2}+3 \right) \\

& \Rightarrow P\equiv \left( \dfrac{-1}{2},\dfrac{5}{2},1 \right) \\

\end{align}$

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