Question

It is given that $a,b,c$ are mutually perpendicular vectors of equal magnitudes.
Statement I vector $(a+b+c)$ is equally inclined to $a,b$ and $c$.
Statement II If $\alpha ,\beta$ and $\gamma$ are the angles at which $(a+b+c)$ is inclined to $a,b$ and $c$, then $\alpha =\beta =\gamma$.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (A)

If $a,b$ and $c$ are mutually perpendicular vectors, then $a\cdot b=b\cdot c=c\cdot a=0$
Given,
$a\cdot b=b\cdot c=c\cdot a=0$
$(\because a,b,c$ are mutually perpendicular)
It is also given that $|a|=|b|=|c|(\because$ All the vectors have same magnitude. Let vector $(a+b+c)$ be inclined to $a,b$ and $c$ at angles $\alpha ,\beta$ and $\gamma$, respectively.
Then, we have
$\cos \alpha =\frac{(a+b+c)\cdot a}{|a+b+c||a|}=\frac{a\cdot a+b\cdot a+c\cdot a}{|a+b+c||a|}$
$=\frac{a\cdot a+0+0}{|a+b+c||a|}$
$=\frac{|a{|}^2}{|a+b+c||a|}(\because a\cdot b=c\cdot b=0)$
$=\frac{|a|}{|a+b+c|}$
$\cos \beta =\frac{(a+b+c)\cdot b}{|a+b+c|\cdot |b|}=\frac{a\cdot b+b\cdot b+c\cdot b}{|a+b+c||b|}$
$=\frac{b\cdot b0+0}{|a+b+c||b|}=\frac{|b|}{|a+b+c||b|}$
$=\frac{|b|}{|a+b+c|}(a\cdot b=c\cdot b=0)$
$\cos \gamma =\frac{(a+b+c)\cdot c}{|a+b+c||c|}=\frac{a\cdot c+b\cdot c+c\cdot c}{|a+b+c||c|}$
$=\frac{c\cdot c+0+0}{|a+b+c||c|}(\because a\cdot c=b\cdot c=0)$
$=\frac{|c{|}^2}{|a+b+c|}=\frac{|c|}{|a+b+c|}$
Now, as $|a|=|b|=|c|$, therefore, $\cos \alpha =\cos \beta =\cos \gamma$
$\therefore \alpha =\beta =\gamma$
Hence, the vector $(a+b+c)$ is equally inclined to $a,b$ and $c$.