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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.5 question 4 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.5 question 4 maths

Answers (1)

Answer:

The required vector equation of plane is a(x - 1) + b(y - 1) + c (z + 1)=0,

Where 5a + 3b - 4c = 0

Hint:

 Convert to the vector form by the given points.

Given:

 Vector equation of plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3)

Solution:

Let A (1, 1, -1), B (6, 4, -5) and C (-4, -2, 3)

The required plane passes through the point A (1, 1, -1) whose vector is

\vec{a}=\hat{i}+\hat{j}-\hat{k}

 and is normal to the vector \vec{n} given by

\begin{aligned} &\vec{n}=\vec{AB}\times \vec{AC}\\ &\vec{AB}=\vec{OB}\times \vec{OA}=(6\hat{i}+4\hat{j}-5\hat{k})-(\hat{i}+\hat{j}-\hat{k})=5\hat{i}+3\hat{j}-4\hat{k} \\ &\vec{AC}=\vec{OC}\times \vec{OA}=(-4\hat{i}-2\hat{j}+3\hat{k})-(\hat{i}+\hat{j}-\hat{k})=-5\hat{i}-3\hat{j}+4\hat{k} \\ &\vec{n}=\vec{AB}\times \vec{AC}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 5 &3 &-4 \\ -5 &-3 &4 \end{vmatrix}=0\hat{i}+0\hat{j}+0\hat{k}=\vec{0} \end{aligned}

So, the given points are collinear.

Thus, there will be an infinite number of planes passing through these points.

Their equations passing through (1, 1, -1) are given by

 a(x - 1) + b(y - 1) + c (z + 1)=0            .....(i)

Since this passes through B (6, 4, -5)

a(6 - 1) + b(4 - 1) + c (-5 + 1)=0

5a + 3b - 4c = 0             .....(ii)

From (i) and (ii) the equations of the infinite planes are

a(x - 1) + b(y - 1) + c (z + 1)=0,

Where 5a + 3b - 4c = 0

Posted by

Gurleen Kaur

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