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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 3 · Set 1 · Q35
(a) Find the value of b so that the lines (x -1)/2=(y - b)/3=(z -3)/4 and (x -4)/5=(y -1)/2= z are intersecting lines. Also, find the point of…
(a)
Find the value of $\displaystyle \mathrm{b}$ so that the lines $\displaystyle \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-\mathrm{b}}{3}=\frac{\mathrm{z}-3}{4}$ and $\displaystyle \frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{z}$ are intersecting lines. Also, find the point of intersection of these given lines.
OR
(b)
Find the equations of all the sides of the parallelogram ABCD whose vertices are $\displaystyle \mathrm{A}(4,7,8), \mathrm{B}(2,3,4), \mathrm{C}(-1,-2,1)$ and $\displaystyle \mathrm{D}(1,2,5)$. Also, find the coordinates of the foot of the perpendicular from A to CD.
Marking-scheme solution
(a)
As lines are intersecting, $\displaystyle \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right)=0\right.$
$$\Rightarrow\left|\begin{array}{ccr}
$\displaystyle 3$ & $\displaystyle 1$-\mathrm{b} & -$\displaystyle 3$
$\displaystyle 2$ & $\displaystyle 3$ & $\displaystyle 4$
$\displaystyle 5$ & $\displaystyle 2$ & $\displaystyle 1$
\end{array}\right|=$\displaystyle 0$
$$$\Rightarrow \mathrm{b}=2$
Any point on the line $\displaystyle \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ is
$$($\displaystyle 2$ \lambda+$\displaystyle 1,3$ \lambda+$\displaystyle 2,4$ \lambda+$\displaystyle 3$), \quad \lambda \in \mathrm{R}
$$For the point of intersection, this point must lie on the line
$$\begin{aligned}
& \frac{\mathrm{x}-$\displaystyle 4$}{$\displaystyle 5$}=\frac{\mathrm{y}-$\displaystyle 1$}{$\displaystyle 2$}=\mathrm{z}
& \Rightarrow \frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=$\displaystyle 4$ \lambda+$\displaystyle 3$
\end{aligned}
$$∴ point of intersection is $\displaystyle (-1,-1,-1)$
(b)
Equation of the line $\displaystyle \mathrm{AB}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-7}{4}=\frac{\mathrm{z}-8}{4}$
Equation of the line $\displaystyle \mathrm{BC}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-4}{3}$
Equation of the line CD : $\displaystyle \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}+2}{2}=\frac{\mathrm{z}-1}{2}$
Equation of the line DA : $\displaystyle \frac{\mathrm{x}-4}{3}=\frac{\mathrm{y}-7}{5}=\frac{\mathrm{z}-8}{3}$
Let P be foot of perpendicular from A to CD .
∴ Coordinates of P are ( $\displaystyle \lambda-1,2 \lambda-2,2 \lambda+1$ ) for some $\displaystyle \lambda$
d.r.'s of AP are ( $\displaystyle \lambda-5,2 \lambda-9,2 \lambda-7$ )
since $\displaystyle \mathrm{AP} \perp \mathrm{CD}$
$$\Rightarrow $\displaystyle 1$(\lambda-$\displaystyle 5$)+$\displaystyle 2$($\displaystyle 2$ \lambda-$\displaystyle 9$)+$\displaystyle 2$($\displaystyle 2$ \lambda-$\displaystyle 7$)=$\displaystyle 0$
$$$\Rightarrow 9 \lambda=37 \quad \Rightarrow \lambda=\frac{37}{9}$
∴ Coordinates of P are $\displaystyle \left(\frac{28}{9}, \frac{56}{9}, \frac{83}{9}\right)$
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