Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{x} \hat{\mathbf{i}}+\mathrm{y} \hat{\mathbf{j}}+\mathrm{z} \hat{\mathbf{k}}$ is a unit vector and $\mathrm{x}: \mathrm{y}: \mathrm{z}=\sqrt{3}: 2: 3$, then what is the value of $z ?$
Options:
Solution:
1490 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{4}$
Let $x \hat{i}+y \hat{j}+z \hat{k}$ is a unit vector.
$\therefore \quad x^{2}+y^{2}+z^{2}=1$
Given $x: y: z=\sqrt{3}: 2: 3$
$\Rightarrow x=\sqrt{3} k, y=2 k$ and $z=3 k$
$\therefore \quad(\sqrt{3} k)^{2}+(2 k)^{2}+(3 k)^{2}=1$
$\Rightarrow 3 k^{2}+4 k^{2}+9 k^{2}=1$
$\Rightarrow k^{2}=\frac{1}{16} \Rightarrow k=\frac{1}{4}$
Hence, $z=3 k=3 \times \frac{1}{4}=\frac{3}{4}$
$\therefore \quad x^{2}+y^{2}+z^{2}=1$
Given $x: y: z=\sqrt{3}: 2: 3$
$\Rightarrow x=\sqrt{3} k, y=2 k$ and $z=3 k$
$\therefore \quad(\sqrt{3} k)^{2}+(2 k)^{2}+(3 k)^{2}=1$
$\Rightarrow 3 k^{2}+4 k^{2}+9 k^{2}=1$
$\Rightarrow k^{2}=\frac{1}{16} \Rightarrow k=\frac{1}{4}$
Hence, $z=3 k=3 \times \frac{1}{4}=\frac{3}{4}$
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.