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Find the Image of the Point with Position Vector 3 ^ I + ^ J + 2 ^ K in the Plane → R ⋅ ( 2 ^ I − ^ J + ^ K ) = 4 . - Mathematics

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Question

Find the image of the point with position vector \[3 \hat{i} + \hat{j}  + 2 \hat{k} \]  in the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j}  + \hat{k}  \right) = 4 .\]  Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i}  + \hat{j}  + 2 \hat{k} .\]

 
 
Sum
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Solution

\[ \text{ Let Q be the image of the point P }  (3 \hat{i}  + \hat{j} +2 \hat{k}  ) \text{ in the plane }  \vec{r .} \left( 2 \hat{i} - \hat{j}  + \hat{k} \right)= 4\]

\[\text{ Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector } 2 \hat{i} - \hat{j} + \hat{k} . \text{ So, the equation of PQ is } \]

\[ \vec{r} = \left( 3 \hat{i} + \hat{j} +2 \hat{k}  \right) + \lambda \left( 2 \hat{i}  - \hat{j}  + \hat{k}  \right)\]

\[\text{ As Q lies on PQ, let the position vector of Q be} \left( 3 + 2\lambda \right) \hat{i}  +\left( 1 - \lambda \right) \hat{j}  +\left( 2 + \lambda \right) \hat{k}  .\]

\[\text{ Let R be the mid-point of PQ. Then, the position vector of R is} \]

\[\frac{\left[ \left( 3 + 2\lambda \right) \hat{i}  +\left( 1 - \lambda \right) \hat{j}  +\left( 2 + \lambda \right) \hat{k } \right]+\left[ 3 \hat{i}  + \hat{j} +2 \hat{k} \right]}{2}\]

\[=\frac{\left( 6 + 2\lambda \right) \hat{i} + \left( 2 - \lambda \right) \hat{j} + \left( 4 + \lambda \right) \hat{k} }{2}\]

\[ = \left( 3 + \lambda \right) \hat{i} + \left( 1 - \frac{\lambda}{2} \right) \hat{j} + \left( 2 + \frac{\lambda}{2} \right) \hat{k}  \]

\[ \text{ Since R lies in the plane }  \vec{r .} \left( 2 \hat{i}  - \hat{j} + \hat{k}  \right)= 4,\]

\[\left[ \left( 3 + \lambda \right) \hat{i}  + \left( 1 - \frac{\lambda}{2} \right) \hat{j}  + \left( 2 + \frac{\lambda}{2} \right) \hat{k}  \right] . \left( 2 \hat{i}  - \hat{j}  + \hat{k}  \right)= 4\]

\[ \Rightarrow 6 + 2\lambda - 1 + \frac{\lambda}{2} + 2 + \frac{\lambda}{2} = 4\]

\[ \Rightarrow 7 + 2\lambda + \frac{\lambda}{2} + \frac{\lambda}{2} = 4\]

\[ \Rightarrow 14 + 6 \lambda = 8\]

\[ \Rightarrow 6 \lambda = 8 - 14\]

\[ \Rightarrow \lambda = - 1\]

\[\text{ Putting } \lambda = - 1\text{  in Q, we get }  \]

\[Q = \left( 3 + 2( - 1) \right) \hat{i}  +\left( 1 - ( - 1) \right) \hat{j}  +\left( 2 + ( - 1) \right) \hat{k}  \]

\[ = \hat{i}  + 2 \hat{j}  + \hat{k}  \text{ or } (1, 2, 1)\]

\[\text{ Therefore, by putting } \lambda = - \text{ 1 in R, we get} \]

\[R = \left( 3 + ( - 1) \right) \hat{i} + \left( 1 - \frac{( - 1)}{2} \right) \hat{j}  + \left( 2 + \frac{( - 1)}{2} \right) \hat{k}  \]

\[ = 2 \hat{i}  + \frac{3}{2} \hat{j}  + \frac{3}{2} \hat{k}  \]

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Chapter 29: The Plane - Exercise 29.15 [Page 81]

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RD Sharma Mathematics [English] Class 12
Chapter 29 The Plane
Exercise 29.15 | Q 4 | Page 81

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