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  • A line 'l' passing through origin is perpendicular to the lines <img alt="l_1:\overrightarrow{r} =(3+t)\widehat{i} +(-1+2t)\widehat{j}+ (4+2t)\widehat{k}" src="https://entrancecorner.codecogs.com/gif.latex?l_1%3A%5Coverrightarrow%7B

A line 'l' passing through origin is perpendicular to the lines
l_1:\overrightarrow{r} =(3+t)\widehat{i} +(-1+2t)\widehat{j}+ (4+2t)\widehat{k}
l_2:\overrightarrow{r} =(3+2s)\widehat{i} +(3+2s)\widehat{j}+ (2+s)\widehat{k}
If the co-ordinates of the point in the first octant on 'l2' at a distance of \sqrt{17} from the point of intersection of 'l' and 'l1' are (a, b, c), then 18(a + b + c) is equal to ______.
Option: 1 36
Option: 2 54
Option: 3 72
Option: 4 44

Answers (1)

best_answer

Let equation of line l is

l: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

since the line, l pass via origin so,

l: \frac{x-0}{a}=\frac{y-0}{b}=\frac{z-0}{c} 

\\\ell_{1}: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}}\\ \ell_{2}: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{\mathrm{i}}+(3+2 \mathrm{~s}) \hat{\mathrm{j}}+(4+\mathrm{s}) \hat{\mathrm{k}}\\

or

\\\ell_{1}: \overrightarrow{\mathrm{r}}=(3\hat i-\hat j+4\hat k)+t(\hat i+2\hat j+2\hat k)\\\ell_{2}: \overrightarrow{\mathrm{r}}=(3\hat i+3\hat j+4\hat k)+s(2\hat i+2\hat j+\hat k)

And also line l is perpendicular to l1 and 12

\\\mathrm{DR} \text { of } \ell_{1} \equiv(1,2,2)\\ \mathrm{DR} \text { of } \ell_{2} \equiv(2,2,1)

\\D R^{\prime} \text { s of } l \text { is }\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array}\right|=-2 \hat{i}+3 \hat{j}+2 \hat{k}

\\l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{2}=k_1\\\text { Now, } A\left(-2 k_{1}, 3 k_{1},-2 k_{1}\right)\\

Point A lies on lie on the line l1

\\\therefore\left(-2 k_{1}\right) \hat{i}+\left(3 k_{1}\right) \hat{j}-\left(2 k_{1}\right) \hat{k}=(3+t) \hat{i}+(-1+2 t) \hat{j} +(4+2 t) \hat{k} \\ \Rightarrow \quad 3+t=-2 k_{1},-1+2 t=3 k_{1}, 4+2 t=-2 k_{1}\\\Rightarrow k_1=-1\\A(2,-3,2)

\\\text{Let any point on } l_{2}\;\;\text{is }(3+2 s, 3+2 s, 2+s)\\ \sqrt{(2-3-2 s)^{2}(-3-3+2 s)^{2}+(2-2-s)^{2}}=\sqrt{17} \\9 s^{2}+28 s+37=17 \\9 s^{2}+28 s+20=0 \\\Rightarrow \quad 9 s^{2}+18 s+10 s+20=0 \\\Rightarrow (9 s+10)(s+2)=0 \\\therefore s=-2, \frac{-10}{9}

\text{ Hence, }(-1,-1,0) \text{ and } \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right) \text{are required points.}

Posted by

himanshu.meshram

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