Question

If $\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{c}}\,$are three mutually perpendicular vectors, then the vector which is equally inclined to three vectors are
(a) $\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\,$
(b) $\dfrac{\overset{\to }{\mathop{a}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|}+\dfrac{\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}+\dfrac{\overset{\to }{\mathop{c}}\,}{\left| \overset{\to }{\mathop{c}}\, \right|}$
(c) $\dfrac{\overset{\to }{\mathop{a}}\,}{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}}+\dfrac{\overset{\to }{\mathop{b}}\,}{{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}}+\dfrac{\overset{\to }{\mathop{c}}\,}{{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}}$
(d) $\left| \overset{\to }{\mathop{a}}\, \right|\overset{\to }{\mathop{a}}\,-\left| \overset{\to }{\mathop{b}}\, \right|\overset{\to }{\mathop{b}}\,+\left| \overset{\to }{\mathop{c}}\, \right|\overset{\to }{\mathop{c}}\,$

Answer 1

Find the relation between vectors $\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{c}}\,$ in case of being mutually perpendicular. Assume $\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|$ is equal to a constant. Find the sum of square of $\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,$and$\overset{\to }{\mathop{c}}\,$. Use the formula to find the angle between vector$\overset{\to }{\mathop{a}}\,$and $\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$, $\overset{\to }{\mathop{b}}\,$and $\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$and$\overset{\to }{\mathop{c}}\,$and$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$. Find the vector which is equally inclined.

Complete step by step answer:
Given to us are the three vectors $\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{c}}\,$which are mutually perpendicular to each other.
Two vectors $\overset{\to }{\mathop{a}}\,$and $\overset{\to }{\mathop{b}}\,$whose dot product $\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=0$are said to be orthogonal. Therefore, the vectors $\overset{\to }{\mathop{a}}\,$and $\overset{\to }{\mathop{b}}\,$are mutually perpendicular. Similarly, dot product of vector $\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{c}}\,$, is 0, and so, they are mutually perpendicular.

& \Rightarrow \overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{.b}}\,=0,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{.c}}\,=0,\overset{\to }{\mathop{c}}\,\overset{\to }{\mathop{.a}}\,=0 \\\ & \Rightarrow \overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{.b}}\,=\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{.c}}\,=\overset{\to }{\mathop{c}}\,\overset{\to }{\mathop{.a}}\,=0-(1) \\\ \end{aligned}$$ Now, let us consider that$$\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=\lambda $$, where $$\lambda $$is a constant. We know the formula, $${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$$. $$\therefore $$The value of $${{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}$$will be, $${{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)$$ We know from equation (1) that$$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\,=0$$. $$\begin{aligned} & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( 0+0+0 \right) \\\ & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\times 0 \\\ & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}} \\\ \end{aligned}$$ Let us substitute $$\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=\lambda $$on the RHS of the expression. $$\begin{aligned} & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\lambda }^{2}}+{{\lambda }^{2}}+{{\lambda }^{2}} \\\ & {{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}=3{{\lambda }^{2}} \\\ \end{aligned}$$ Taking square root on both sides, $$\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|=\sqrt{3}\lambda $$. Now, let us find the angle between $$\overset{\to }{\mathop{a}}\,$$and $$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$. The formula for find the angle between the two vectors is, $$\cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|.\left| \overset{\to }{\mathop{b}}\, \right|}$$. The cosine of the angle between 2 vectors is equal to the dot product of this vector divided by the product of vector magnitude. Here, let us take the two vectors as $$\overset{\to }{\mathop{a}}\,$$and$$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$. $$\therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)}{\left| \overset{\to }{\mathop{a}}\, \right|.\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}$$ Let us take dot product in the numerator. Substitute the value of $$\overset{\to }{\mathop{a}}\,$$and $$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$in the denominator. $$\therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\,}{\lambda \times \sqrt{3}\lambda }$$ We know, $$\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\,=0$$and $${{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}={{\lambda }^{2}}$$. $$\begin{aligned} & \therefore \cos \alpha =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}} \\\ & \therefore \alpha ={{\cos }^{-1}}\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ Similarly, we can find the angle between $$\overset{\to }{\mathop{b}}\,$$and$$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$.$$\begin{aligned} & \therefore \cos \alpha =\dfrac{\overset{\to }{\mathop{b}}\,.\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)}{\left| \overset{\to }{\mathop{b}}\, \right|.\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}=\dfrac{\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,}{\lambda \times \sqrt{3}\lambda } \\\ & =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ We know, $$\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{a}}\,=0,\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,=0$$, so above equation can be written as, $$\therefore \cos \alpha =\dfrac{{{\lambda }^{2}}}{\sqrt{3}{{\lambda }^{2}}}=\dfrac{1}{\sqrt{3}}$$ $$\therefore \alpha ={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$$ Similarly, the angle between $$\overset{\to }{\mathop{c}}\,$$and $$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$is$${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$$. Hence, angle between $$\overset{\to }{\mathop{a}}\,$$and $$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$=$${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$$. Angle between $$\overset{\to }{\mathop{b}}\,$$and$$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$=$${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$$. Angle between $$\overset{\to }{\mathop{c}}\,$$and $$\left( \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right)$$=$${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$$. $$\therefore \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\,$$is equally inclined with $$\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{c}}\,$$. **So, the correct answer is “Option A”.** **Note:** The dot product between 2 unit vectors and itself is simple to compute. If the vectors are of length one, then the dot product of three vectors $$\overset{\to }{\mathop{i}}\,,\overset{\to }{\mathop{j}}\,$$and $$\overset{\to }{\mathop{k}}\,$$becomes, $$\overset{\to }{\mathop{i}}\,.\overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,.\overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,.\overset{\to }{\mathop{k}}\,=1$$. If the vectors are of length a, then the dot product becomes, $$\overset{\to }{\mathop{i}}\,.\overset{\to }{\mathop{i}}\,=\overset{\to }{\mathop{j}}\,.\overset{\to }{\mathop{j}}\,=\overset{\to }{\mathop{k}}\,.\overset{\to }{\mathop{k}}\,=a$$