Search any question & find its solution
Question:
Answered & Verified by Expert
If the position vectors of the vertices of $\triangle A B C$ are $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $5(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$, respectively. Then, the magnitude of the altitude from $A$ onto the side $B C$ is
Options:
Solution:
2532 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{3} \sqrt{5}$
Let altitude of a triangle be $p$.
Now,
$\begin{aligned}
& \mathrm{AB}=-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \mathrm{AC}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\
& \mathrm{BC}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}
\end{aligned}$
We know that,
Area of $\triangle A B C=\frac{1}{2}|\mathrm{AB} \times \mathrm{AC}|=\frac{1}{2}|\mathrm{BC}| \times p$
Now, $\mathrm{AB} \times \mathrm{AC}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & -1 & 2 \\ 2 & 1 & 6\end{array}\right|=-8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}$
From Eq. (i),
$\begin{gathered}
\frac{1}{2} \sqrt{(-8)^2+(16)^2+(0)^2}=\frac{1}{2} \sqrt{4^2+2^2+4^2} \times p \\
\Rightarrow \quad p=\frac{\sqrt{320}}{6}=\frac{8 \sqrt{5}}{6}=\frac{4 \sqrt{5}}{3}
\end{gathered}$
Now,
$\begin{aligned}
& \mathrm{AB}=-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \mathrm{AC}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\
& \mathrm{BC}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}
\end{aligned}$
We know that,
Area of $\triangle A B C=\frac{1}{2}|\mathrm{AB} \times \mathrm{AC}|=\frac{1}{2}|\mathrm{BC}| \times p$
Now, $\mathrm{AB} \times \mathrm{AC}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & -1 & 2 \\ 2 & 1 & 6\end{array}\right|=-8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}+0 \hat{\mathbf{k}}$
From Eq. (i),
$\begin{gathered}
\frac{1}{2} \sqrt{(-8)^2+(16)^2+(0)^2}=\frac{1}{2} \sqrt{4^2+2^2+4^2} \times p \\
\Rightarrow \quad p=\frac{\sqrt{320}}{6}=\frac{8 \sqrt{5}}{6}=\frac{4 \sqrt{5}}{3}
\end{gathered}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.