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|} $
$=\frac{b \cdot b 0+0}{|a+b+c||b|}=\frac{|b|}{|a+b+c||b|}$
$ =\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 $c$.