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  • Let the position vectors of two points P and Q be and <img alt="\hat{i}+2\hat{j}-4\hat{k}," src="/latex-image/?%5Chat%7Bi%7D&plus;2

Let the position vectors of two points P and Q be 3 \hat{i}-\hat{j}+2\hat{k} and \hat{i}+2\hat{j}-4\hat{k}, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are ( 4, -1, 2 ) and ( -2, 1, -2 ), respectively. Let lines PR and QS intersect at T. If the vector \overrightarrow{TA} is perpendicular to both \overrightarrow{PR} and \overrightarrow{QS} and the length of vector \overrightarrow{TA} is \sqrt{5} units, then the modulus of a position vector of A is :
Option: 1 \sqrt{5}
Option: 2 \sqrt{482}
Option: 3 \sqrt{227}
Option: 4 \sqrt{171}

Answers (1)

best_answer

 

\begin{aligned} &\mathrm{P}(3,-1,2) \\ &\mathrm{Q}(1,2,-4) \\ &\overrightarrow{\mathrm{PR}} \| 4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ &\overrightarrow{\mathrm{QS}} \|-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \end{array}

dr's of normal to the plane containing P, T and Q will be proportional to :

\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & -1 & 2 \\ -2 & 1 & -2 \end{array}\right|

\begin{aligned} &\therefore \quad \frac{\ell}{0}=\frac{\mathrm{m}}{4}=\frac{\mathrm{n}}{2} \\ &\text { For point, } \mathrm{T}: \overrightarrow{\mathrm{PT}}=\frac{\mathrm{x}-3}{4}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\lambda \\ &\overrightarrow{\mathrm{QT}}=\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}+4}{-2}=\mu \end{array}

\begin{aligned} &\mathrm{T}:(4 \lambda+3,-\lambda-1,2 \lambda+2) \quad \cong(2 \mu+1, \mu+2,-2 \mu-4) \\& 4 \lambda+3=-2 \mu+1 \quad \Rightarrow 2 \lambda+\mu=-1 \\ &\lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2 \\ &\& \quad\mu=-5 \quad \lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2 \\& \text {So point } T:(11,-3,6) \end{array}

\begin{aligned} &\overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm\left(\frac{2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{5}}\right) \sqrt{5} \\ &\overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm(2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ &\overrightarrow{\mathrm{OA}}=11 \hat{\mathrm{i}}-\hat{\mathrm{j}}+7 \hat{\mathrm{k}} \end{array}

\begin{aligned} &\text {Or, }\\&9 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\\ &|\overrightarrow{\mathrm{OA}}|=\sqrt{121+1+49}=\sqrt{171}\\ &\text { or }\\ &\sqrt{81+25+25}=\sqrt{131} \end{aligned}

Posted by

Kuldeep Maurya

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