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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 1 · Q34
(a) Find the co-ordinates of the foot of the perpendicular drawn from the point (2,3,-8) to the line (4-x)/2=y/6=(1- z)/3. Also, find the…
Find the co-ordinates of the foot of the perpendicular drawn from the point $\displaystyle (2,3,-8)$ to the line $\displaystyle \frac{4-x}{2}=\frac{\mathrm{y}}{6}=\frac{1-\mathrm{z}}{3}$. Also, find the perpendicular distance of the given point from the line.
Find the shortest distance between the lines $\displaystyle \mathrm{L}_{1} \& \mathrm{~L}_{2}$ given below : $\displaystyle \mathrm{L}_{1}$ : The line passing through $\displaystyle (2,-1,1)$ and parallel to $\displaystyle \frac{x}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{3}$ \[\mathrm{L}_{2}: \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+(2 \mu+1) \hat{\mathrm{j}}-(\mu+2) \hat{\mathrm{k}} \]
Marking-scheme solution
(a)
The standard form of the equation of the line is $\displaystyle \frac{x-4}{-2}=\frac{\mathrm{y}}{6}=\frac{\mathrm{z}-1}{-3}$
Let foot of the perpendicular from the point $\displaystyle \mathbf{A}(\mathbf{2}, 3,-\mathbf{8})$ to the given line be $\displaystyle B(-2 \lambda+4,6 \lambda,-3 \lambda+1)$
D-ratios of AB is: $\displaystyle -2 \lambda+2,6 \lambda-3,-3 \lambda+9$
As AB is perpendicular to the given line: $\displaystyle -2(-2 \lambda+2)+6(6 \lambda-3)-3(-3 \lambda+9)=0$
$$\Rightarrow \lambda=$\displaystyle 1$
$$∴ Foot of the perpendicular is: $\displaystyle \mathbf{B}(\mathbf{2 , 6 , - 2})$
$$\text { Perpendicular distance }=\mathrm{AB}=$\displaystyle 3$ \sqrt{5}
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.