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  • Let a line l pass through the origin and be perpendicular to the lines <img alt="\begin{aligned} & 1_1: \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}+

Let a line l pass through the origin and be perpendicular to the lines \begin{aligned} & 1_1: \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}+\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \lambda \in \mathbb{R} \text { and } \\ & 1_2: \overrightarrow{\mathrm{r}}=-\hat{\mathrm{i}}+\hat{\mathrm{k}}+\mu 2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \mu \in \mathbb{R} . \end{aligned}

If P is the point of intersection of l and l1, and Q (\alpha, \beta ,\gamma) is the foot of perpendicular from P on l2 , then
9 (\alpha+ \beta+\gamma) is equal to _____:

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Let } \ell=(0 \hat{i}+0 \hat{j}+0 \hat{k})+\gamma(a \hat{i}+b \hat{j}+c \hat{k}) \\ & =\gamma(a \hat{i}+b \hat{j}+c \hat{k}) \\ & \text { a } \hat{\mathrm{i}}+\hat{\mathrm{b}}+\mathrm{c} \hat{\mathrm{k}}=\left|\begin{array}{lll} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{array}\right| \\ & =\hat{\mathrm{i}}(2-6)-\hat{\mathrm{j}}(1-6)+\hat{\mathrm{k}}(2-4) \\ & =-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \ell=\gamma(-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \end{aligned} \mathrm{P}$ is intersection of $\ell$ and $\ell_1 -4 \gamma=1+\lambda, 5 \gamma=-11+2 \lambda,-2 \gamma=-7+3 \lambda

By solving these equation \gamma=-1, \mathrm{P}(4,-5,2) Let \mathrm{Q}(-1+2 \mu, 2 \mu, 1+\mu) \begin{aligned} & \overrightarrow{\mathrm{PQ}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=0 \\ & -2+4 \mu+4 \mu+1+\mu=0 \\ & 9 \mu=1 \end{aligned} \mu=\frac{1}{9} \begin{aligned} & Q\left(\frac{-7}{9}, \frac{2}{9}, \frac{10}{9}\right) \\ & 9(\alpha+\beta+\gamma)=9\left(\frac{-7}{9}+\frac{2}{9}+\frac{10}{9}\right) \\ & =5 \end{aligned}

Posted by

Ritika Harsh

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