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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 6 Maths Textbook Solution.

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Answer: L.H.S = R.H.S

Hint: use vector set product.

Given\vec{r}=\left ( i+2j-k \right )=3

Solution: we know that plane \vec{r}.\vec{n}=d  contains the line \begin{aligned} &\pi=a+\lambda b+i j \\ &\vec{b} \cdot \vec{n}=0 \\ &\vec{a} \cdot \vec{n}=d \end{aligned}

Given equation of plane;

                \bar{\pi }\left ( i+2j-k \right )=3and equation of line  \bar{\pi }=i+j+\lambda \left ( 2i+j+4k \right )

                So,\bar{n}=i+2j-k,\bar{a}=i+j

                \begin{aligned} &\overline{3 \bar{b}}=2 i+j+4 k \\ &\vec{b} \cdot \vec{n}=(2 i+j 4 k)(i+2 j-k) \\ &=2+2-4 \\ &\vec{b} \cdot \vec{n}=0 \\ &\vec{a} \cdot \vec{n}=(i+j)(i+2 j-k) \\ &=1+2-0 \\ &=3 \\ &=\mathrm{d} \end{aligned}

Since \vec{b}.\vec{n}=0  and \vec{a}.\vec{n}=d  so the line is the given plane hence proved

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