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Home » Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.
Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Three Dimensional Geometry
Find the length and the foot of perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z + 5 = 0.

 

According to ques,

plane is:

    \[2x\text{ }\text{ }2y\text{ }+\text{ }4z\text{ }+\text{ }5\text{ }=\text{ }0\]

and point:

    \[\left( 1,\text{ }3/2,\text{ }2 \right)\]

The direction ratios of the normal to the plane are:

    \[2,\text{ }-2,\text{ }4\]

Hence,

Equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is:

NCERT Exemplar Solutions Class 12 Mathematics Chapter 11 - 29

Since any point in the plane is:

    \[2\lambda \text{ }+\text{ }1,\text{ }-2\lambda \text{ }+\text{ }3/2,\text{ }4\lambda \text{ }+\text{ }2\]

Also, the point lies in the plane, then

    \[2\left( 2\lambda \text{ }+\text{ }1 \right)\text{ }\text{ }2\left( -2\lambda \text{ }+\text{ }3/2 \right)\text{ }+\text{ }4\left( 4\lambda \text{ }+\text{ }2 \right)\text{ }+\text{ }5\text{ }=\text{ }0\]

Or,

    \[4\lambda \text{ }+\text{ }2\text{ }+\text{ }4\lambda \text{ }\text{ }3\text{ }+\text{ }16\lambda \text{ }+\text{ }8\text{ }+\text{ }5\text{ }=\text{ }0\]

Or,

    \[24\lambda \text{ }+\text{ }12\text{ }=\text{ }0\text{ }\lambda \text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

Hence, the coordinates of the point in the plane are:

    \[2\left( -1/2 \right)\text{ }+\text{ }1,\text{ }-2\left( -1/2 \right)\text{ }+\text{ }3/2,\text{ }4\left( -1/2 \right)\text{ }+\text{ }2\text{ }=\text{ }0,\text{ }5/2,\text{ }0\]

Therefore, the foot of the perpendicular is (0, 5/2, 0) and the required length

NCERT Exemplar Solutions Class 12 Mathematics Chapter 11 - 30

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