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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 16 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 16 maths

Answers (1)

Answer:

 -\frac{13}{4}

Hint:

 Use this formula if two planes are perpendicular a1a2+b1b2+c1c2=0

Given:

 \frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k} \perp \text { to normal of plane } \overrightarrow{r}.(2\widehat{i}+3\widehat{j}+4\widehat{k})=4

Solution:

\overrightarrow{r}.(2\widehat{i}+3\widehat{j}+4\widehat{k})=4 \qquad \qquad \dots(i)

So the vector normal to the plane as,

\overrightarrow{n}=2\widehat{i}+3\widehat{j}+4\widehat{k}

a1 = 2, b1 = 3, c1 = 4

and

\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k} \qquad \qquad \dots (ii)

a2 = 2, b2 = 3, c2 = k

Since (i) and (ii) ⊥ to each other.

\begin{aligned} &a_1a_2+b_1b_2+c_1C_2=0\\ &2.2+3.3+4.k=0\\ &\Rightarrow 4+9+4k=0\\ &\Rightarrow 4k=-13\\ &\Rightarrow k=-\frac{13}{4} \end{aligned}

Posted by

Gurleen Kaur

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