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Equation of a plane perpendicular to a given vector and passing through a given point - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Equation of a plane perpendicular to a given vector and passing through a given point is considered one of the most asked concept.

  • 22 Questions around this concept.

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QuestionMEDIUM

The coordinates of the foot of the perpendicular from the point (1, −2, 1) on the plane containing the lines

\small \frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}  and

\small \frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7} , is

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A vector \overrightarrow{\mathrm{v}} in the first octant is inclined to the x- axis at 60^{\circ} , to the y-axis at 45 and to the z-axis at an acute angle. If a plane passing through the points (\sqrt{2},-1,1) and (a, b, c), is normal to \vec{v},then

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If the equation of the plane containing the line x+2 y+3 z-4=0\, 2 x+y-z+5 and perpendicular to the plane \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) is  a x+b y+c z=4, then (a-b+c) is equal to:

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If the equation of the plane that contains the point (-2,3,5) and is perpendicular to each of the planes(2x+4y+5z)=8 and 3x-2y-3z=5 is \alpha x+\beta y+\gamma z+97=0 then \alpha +\beta +\gamma =

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The plane, passing through the points (0, -1, 2) and (-1, 2, 1) and parallel to the line passing through (5,1,-7) and (1,-1,-1), also passes through the point:

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Concepts Covered - 1

Equation of a plane perpendicular to a given vector and passing through a given point

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. 

Let \vec{\mathbf n} = a\hat {\mathbf i}+b\hat {\mathbf j}+c\hat {\mathbf k} be a vector and P (x0, y0, z0) be a point. Then the set of all point Q (x, y, z) such that \overrightarrow{PQ} orthogonal to \vec{\mathbf n} forms a plane.

We say that \vec{\mathbf n} is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane:

\vec{\mathbf n}\cdot \overrightarrow{PQ}=0

If position vector of point P is \vec{\mathbf p} and position vector of point Q is \vec{\mathbf q}, then 

\left ( \vec{\mathbf q}-\vec{\mathbf p} \right )\cdot \vec{\mathbf n}=\mathbf 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\text{As }\overrightarrow{PQ}=\vec{\mathbf q}-\vec{\mathbf p})

This is the vector equation of the plane.


 

Cartesian form

\\\text{Position vector of point P and point Q is }\;\vec{\mathbf p}=x_0\hat i+y_0\hat j+z_0\hat k\;\text{and }\\\;\vec{\mathbf q}=x\hat i+y\hat j+z\hat k\;\;\text{respectively and vector }\vec{\mathbf n}\;\;\text{is} \;\;a\hat i+b\hat j+c\hat k \\\text{Then,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left ( \vec{\mathbf q}-\vec{\mathbf p} \right )\cdot \vec{\mathbf n}=\mathbf 0\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;}\left ( \left ( x\hat i+y\hat j+z\hat k \right )-\left ( x_0\hat i+y_0\hat j+z_0\hat k \right ) \right )\cdot \left ( a\hat i+b\hat j+c\hat k \right )=0\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left[\left(x-x_{0}\right) \hat{i}+\left(y-y_{0}\right) \hat{j}+\left(z-z_{0}\right) \hat{k}\right] \cdot(\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k})=0\\\mathrm{i.e.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathbf{a}\left(x-x_{0}\right)+\mathbf{b}\left(y-y_{0}\right)+\mathbf{c}\left(z-z_{0}\right)=\mathbf{0}

Thus, the coefficients of x, y and z in the cartesian equation of a plane are the direction ratios of the normal to the plane.

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Equation of a plane perpendicular to a given vector and passing through a given point

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Equation of a plane perpendicular to a given vector and passing through a given point

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 4.28

Line : 6

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