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  • Let L be the line of intersection of planes <img alt="\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=2 \quad$ and $\quad \overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=

Let L be the line of intersection of planes \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=2 \quad$ and $\quad \overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=2$. If $\mathrm{P}(\alpha, \beta, \gamma) is the foot of perpendicular on \mathrm{L} from the point (1,2,0), then the value of 35(\alpha+\beta+\gamma) is equal to :
Option: 1 101
Option: 2 119
Option: 3 143
Option: 4 134

Answers (1)

best_answer

Equation of planes are
P_{1}: \quad x-y+2 z-2=0$ with $\overrightarrow{n_{1}}=i-j+2 k.

P_{2}: \quad 2 x+y-z-2=0$ with $\vec{n}_{2}=2 i+j-k.

Let

P will lie on both planes.

\begin{aligned} &\alpha-\beta+2 \gamma-2=0 \\ \end{aligned} \text { (i) }  and

\begin{aligned} &2 \alpha+\beta-\gamma-2=0 \end{aligned} \text { (ii) }

Also \overrightarrow{P Q}$ will be $\perp$ to line $L.
\Rightarrow \overrightarrow{P Q}$ will be $\perp$ to $\vec{n}_{1} \times \vec{n}_{2}
\begin{aligned} &\Rightarrow \overrightarrow{P Q} \cdot\left(\vec{n_1} \times \overrightarrow{n_{2}}\right)=0 \\ &\Rightarrow((\alpha-1) i+(\beta-2) j+\gamma k) \cdot(-i+5 j+3 k)=0 \\ &\Rightarrow 1-\alpha+5 \beta-10+3 \gamma=0 \\ &\Rightarrow \quad \alpha-5 \beta-3 \gamma+9=0 \quad \text { (iii) } \end{aligned}

Solving (i), (ii) and (iii)

\alpha=\frac{99}{105}, \beta=\frac{135}{105}, \gamma=\frac{123}{105} .

35(\alpha+\beta+\gamma)=119

Hence option (2) is correct.

Posted by

Kuldeep Maurya

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