Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position
vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \hat{\beta} \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$
Options:
vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \hat{\beta} \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$
Solution:
1590 Upvotes
Verified Answer
The correct answer is:
form an equilateral triangle
$\alpha, \beta$ and $\gamma$ be distinct real numbers $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=\beta \hat{\mathrm{i}}+\gamma \hat{\mathrm{j}}+\alpha \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=\gamma \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$
Let $\alpha=1, \beta=2$ and $\gamma=3$
then, $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
Now, $|\overrightarrow{a b}|=\sqrt{(2-1)^{2}}+(3-2)^{2}+(1-3)^{2}=\sqrt{6}$
Similarly $|\overrightarrow{\mathrm{bc}}|=|\overrightarrow{\mathrm{ac}}|=\sqrt{6}$
Hence, point for equilateral triangle.
$\overrightarrow{\mathrm{b}}=\beta \hat{\mathrm{i}}+\gamma \hat{\mathrm{j}}+\alpha \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=\gamma \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$
Let $\alpha=1, \beta=2$ and $\gamma=3$
then, $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
Now, $|\overrightarrow{a b}|=\sqrt{(2-1)^{2}}+(3-2)^{2}+(1-3)^{2}=\sqrt{6}$
Similarly $|\overrightarrow{\mathrm{bc}}|=|\overrightarrow{\mathrm{ac}}|=\sqrt{6}$
Hence, point for equilateral triangle.
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.