Answer
Verified
477.6k+ 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
$\begin{align}
& P\equiv \left( 3\times \dfrac{-1}{2}+1,-\dfrac{-1}{2}+2,4\times \dfrac{-1}{2}+3 \right) \\
& \Rightarrow P\equiv \left( \dfrac{-1}{2},\dfrac{5}{2},1 \right) \\
\end{align}$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
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
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE