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

Answer

Verified

360k+ views

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}$

Last updated date: 25th Sep 2023

â€¢

Total views: 360k

â€¢

Views today: 6.60k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE