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Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise very short answer type question 15

Answers (1)

Answer:

 \overrightarrow{r}=\overrightarrow{a}+\left ( \frac{\overrightarrow{a}-\overrightarrow{n}}{\overrightarrow{b}-\overrightarrow{n}} \right ).\overrightarrow{b}

Hint:

 Put Eqn (i) in Eqn (ii)

Given:

 \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \text { meets the plane }\overrightarrow{r}.\overrightarrow{n}=0

Solution:

\begin{aligned} &\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \qquad \qquad \dots(i)\\ &\overrightarrow{r}.\overrightarrow{n}=0 \qquad \qquad \dots(ii) \end{aligned}

Put (i) in(ii).

\begin{aligned} &(\overrightarrow{a}+\lambda \overrightarrow{b}).\overrightarrow{n}=0\\ &\Rightarrow \overrightarrow{a}.\overrightarrow{n}+\lambda \overrightarrow{b}.\overrightarrow{n}=0\\ &\Rightarrow \lambda \overrightarrow{b}.\overrightarrow{n}=-\overrightarrow{a}.\overrightarrow{n}\\ &\Rightarrow \lambda =-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}} \end{aligned}

Putting the value of λ in eqn (i), we get

\vec{r}=\vec{a}-(\frac{\vec{a}\cdot\vec{n}}{\vec{b}\cdot\vec{n}})\vec{b}

Posted by

Gurleen Kaur

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