3D Geometry 2 Question 9
####9. A line $l$ passing through the origin is perpendicular to the lines
(2013 Adv.)
$l_{1}:(3+t) \hat{\mathbf{i}}+(-1+2 t) \hat{\mathbf{j}}+(4+2 t) \hat{\mathbf{k}},-\infty<t<\infty$
$l_{2}:(3+2 s) \hat{\mathbf{i}}+(3+2 s) \hat{\mathbf{j}}+(2+s) \hat{\mathbf{k}},-\infty<s<\infty$
Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ is (are)
(a) $\frac{7}{3}, \frac{7}{3}, \frac{5}{3}$
(b) $(-1,-1,0)$
(c) $(1,1,1)$
(d) $\frac{7}{9}, \frac{7}{9}, \frac{8}{9}$
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Answer:
Correct Answer: 9. $(\mathrm{b}, \mathrm{d})$
Solution:
- Key Idea Equation of straight line is $/: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Since, $I$ is perpendicular to $I_{1}$ and $I_{2}$.
So, its DR’s are cross-product of $I_{1}$ and $I_{2}$
Now, to find a point on $I_{2}$ whose distance is given, assume a point and find its distance to obtain point.
Let $l: \frac{x-0}{a}=\frac{y-0}{b}=\frac{z-0}{c}$
which is perpendicular to
$ \begin{aligned} & l_{1}:(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & l_{2}:(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \therefore \quad & \text { DR’s of } l \text { is }\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array}\right|=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=k_{1}, k_{2} \end{aligned} $
Now, $A\left(-2 k_{1}, 3 k_{1},-2 k_{1}\right)$ and $B\left(-2 k_{2}, 3 k_{2},-2 k_{2}\right)$.
Since, $A$ lies on $l_{1}$.
$ \begin{aligned} & \therefore \quad\left(-2 k_{1}\right) \hat{\mathbf{i}}+\left(3 k_{1}\right) \hat{\mathbf{j}}-\left(2 k_{1}\right) \hat{\mathbf{k}}=(3+t) \hat{\mathbf{i}}+(-1+2 t) \hat{\mathbf{j}} \\ & +(4+2 t) \hat{\mathbf{k}} \\ & \Rightarrow \quad 3+t=-2 k_{1},-1+2 t=3 k_{1}, 4+2 t=-2 k_{1} \\ & \therefore \quad k_{1}=-1 \end{aligned} $
$\Rightarrow$
$ A(2,-3,2) $
Let any point on $l_{2}(3+2 s, 3+2 s, 2+s)$
$ \begin{array}{rcrl} & & \sqrt{(2-3-2 s)^{2}+(-3-3-2 s)^{2}+(2-2-s)^{2}}=\sqrt{17} \\ \Rightarrow & & 9 s^{2}+28 s+37=17 \\ \Rightarrow & 9 s^{2}+28 s+20=0 \\ \Rightarrow & 9 s^{2}+18 s+10 s+20=0 \\ \Rightarrow & (9 s+10)(s+2)=0 \\ \therefore & s & s=-2, \frac{-10}{9} . \end{array} $
Hence, $(-1,-1,0)$ and $\frac{7}{9}, \frac{7}{9}, \frac{8}{9}$ are required points.
