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  • If   are two values of <img alt="\lambda" src="https://entrancecorner.o

If \lambda_1<\lambda_2  are two values of \lambda such that the angle between the planes P_1: \overrightarrow{\mathrm{r}} \cdot(3 \hat{\imath}-5 \hat{\jmath}+\hat{\mathrm{k}})=7  andP_2: \overrightarrow{\mathrm{r}} \cdot(\lambda \hat{\imath}+\hat{\jmath}-3 \hat{k})=9 is \sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)  then the square of the length of perpendicular from the point 

\left(38 \lambda_1, 10 \lambda_2, 2\right) to the plane P_{1} is 

Option: 1

315 s


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

Plane 

\begin{aligned} & P_1: \vec{r} \cdot(3 \hat{i}-5 \hat{j}+\hat{k})=7 \\ & P_2: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 k)=9 \end{aligned}

angle between plane is same as angle between their normal.
angle between normal \theta then

\begin{aligned} & \operatorname{Cos} \theta= \frac{\langle 3,-5,1\rangle \cdot\langle\lambda, 1,-3\rangle}{\sqrt{9+25+1} \sqrt{\lambda^2+1+9}} \\ & \operatorname{Cos} \theta=\frac{3 \lambda-5-3}{\sqrt{35} \sqrt{\lambda^2+10}} \\ & \because \quad \theta=\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right) \text { then } \\ & \sin \theta=\frac{2 \sqrt{6}}{5} \\ & \cos \theta=\frac{1}{5} \end{aligned}________________(i)

from equation (1)

\begin{aligned} & \frac{3 \lambda-8}{\sqrt{35} \sqrt{\lambda^2+10}}=\frac{1}{5} \\ & \Rightarrow \frac{(3 \lambda-8)^2}{35\left(\lambda^2+10\right)}=\frac{1}{25} \\ & \Rightarrow 5(3 \lambda-8)^2=7\left(\lambda^2+10\right) \\ & \Rightarrow 5\left(9 \lambda^2-48 \lambda+64\right)=7 \lambda^2+70 \\ & \Rightarrow 38 \lambda^2-240 \lambda+250=0 \\ & \Rightarrow 19 \lambda^2-120 \lambda+125=0 \\ & \Rightarrow \lambda=5, \frac{25}{19} \\ \end{aligned}

\lambda_1=\frac{25}{19}, \lambda_2=5

Point \left(38 \lambda_1, 10 \lambda_2, 2\right) \equiv(50,50,2)

\text { distance of }(50,50,2) \text { from plane } P_1 \text { is }

\begin{aligned} & \mathrm{d}=\left|\frac{3 \times 50-5 \times 50+2-7}{\sqrt{9+25+1}}\right| \\ & \mathrm{d}=\left|\frac{150-250+2-7}{\sqrt{35}}\right| \\ & d=\left|\frac{105}{\sqrt{35}}\right| \\ & d=3 \sqrt{35} \\ & d^2=315 \end{aligned}

Posted by

Sanket Gandhi

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