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Question: How do you find a vector equation for the line through the point \(\left( 5,1,4 \right)\) and perpen...

How do you find a vector equation for the line through the point (5,1,4)\left( 5,1,4 \right) and perpendicular to the plane x−2y+z=1x-2y+z=1 ?

Explanation

Solution

To solve these types of questions we will use the concept of parametric equations to write the vector equation for the line. To write the vector equation of the line, first find the normal to the plane and the line perpendicular to the plane will be parallel to the normal, which will give us our equation.

Complete step by step answer:
It is given to us that:
Equation of the plane: x−2y+z=1x-2y+z=1
Point through which the line passes: (5,1,4)\left( 5,1,4 \right)
We know that the general equation of a plane can be given as a(x−x1)+b(y−y1)+c(z−z1)=0a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0
Now, first we find the normal to the plane given, which can be given by the equation n→ =(a,b,c)\overset{\to }{\mathop{n}}\,=\left( a,b,c \right) where aa , bb and cc are the coefficients of the xx , yy and zz coordinates respectively, in the equation of the plane.
Comparing the given equation of the plane in the question and the general equation of a plane, we get,
a=1a=1 , b=−2b=-2 and c=1c=1
So, we get that the equation for the normal perpendicular to the plane x−2y+z=1x-2y+z=1 will be n→ =(1,−2,1)\overset{\to }{\mathop{n}}\,=\left( 1,-2,1 \right) .
We know that the line perpendicular to the plane x−2y+z=1x-2y+z=1 will be parallel to the normal n→ =(1,−2,1)\overset{\to }{\mathop{n}}\,=\left( 1,-2,1 \right) and will pass through the points (5,1,4)\left( 5,1,4 \right)as given in the question.
Therefore, the equation of the line can be given by r→ =a→ +λn→ \overset{\to }{\mathop{r}}\,=\overset{\to }{\mathop{a}}\,+\lambda \overset{\to }{\mathop{n}}\, , where a→ \overset{\to }{\mathop{a}}\, represents the point through which the line passes.
Substituting the required values in the above equation for the line, we get,
⇒r→ =(5,1,4)+λ(1,−2,1)\Rightarrow \overset{\to }{\mathop{r}}\,=\left( 5,1,4 \right)+\lambda \left( 1,-2,1 \right)

Hence, the vector equation of a line perpendicular to the plane x−2y+z=1x-2y+z=1 and passing through the points (5,1,4)\left( 5,1,4 \right) will be r→ =(5,1,4)+λ(1,−2,1)\overset{\to }{\mathop{r}}\,=\left( 5,1,4 \right)+\lambda \left( 1,-2,1 \right)

Note: While solving these questions, understanding of the basic concepts of lines, planes and normal will help a lot. Also, be careful while substituting the points and the values in the equations, otherwise the vector equations formed will be wrong. Knowledge about the formation of parametric equations will also prove to be beneficial.