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Question: If \(\overrightarrow{a}\) is perpendicular to \[\overrightarrow{b}\] and \[\overrightarrow{r}\] is a...

If a→\overrightarrow{a} is perpendicular to b→\overrightarrow{b} and r→\overrightarrow{r} is a non – zero vector such that p→r→+(r→.b→)a→=c→\overrightarrow{p}\overrightarrow{r}+\left( \overrightarrow{r}.\overrightarrow{b} \right)\overrightarrow{a}=\overrightarrow{c}, then r→\overrightarrow{r} is equal to,
A. c→p→−(b→.c→)a→p2\dfrac{\overrightarrow{c}}{\overrightarrow{p}}-\dfrac{\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}}{{{p}^{2}}}
B. a→p→−(c→.a→)b→p2\dfrac{\overrightarrow{a}}{\overrightarrow{p}}-\dfrac{\left( \overrightarrow{c}.\overrightarrow{a} \right)\overrightarrow{b}}{{{p}^{2}}}
C. b→p→−(a→.b→)c→p2\dfrac{\overrightarrow{b}}{\overrightarrow{p}}-\dfrac{\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}}{{{p}^{2}}}
D. c→p2−(b→.c→)a→p→\dfrac{\overrightarrow{c}}{{{p}^{2}}}-\dfrac{\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}}{\overrightarrow{p}}

Explanation

Solution

Hint:Take a dot product of the given equation with b→\overrightarrow{b}and solve for r→\overrightarrow{r} and (r→.b→)\left( \overrightarrow{r}.\overrightarrow{b} \right) using the two linear equations in two variables.

Complete Step-by-step answer:
Given,
a→⊥b→\overrightarrow{a}\bot \overrightarrow{b},
r→≠0\overrightarrow{r}\ne 0
and p→r→+(r→.b→)a→=c→\overrightarrow{p}\overrightarrow{r}+\left( \overrightarrow{r}.\overrightarrow{b} \right)\overrightarrow{a}=\overrightarrow{c}.
Let us multiply the above equation with b→\overrightarrow{b} to get a scalar product.
p→(r→.b→)+(r→.b→)(a→.b→)=c→.b→ ∵a→⊥b→⇒a→.b→=0 ∴p→(r→.b→)=c→.b→ ⇒(r→.b→)=c→.b→p→ \begin{aligned} & \overrightarrow{p}\left( \overrightarrow{r}.\overrightarrow{b} \right)+\left( \overrightarrow{r}.\overrightarrow{b} \right)\left( \overrightarrow{a}.\overrightarrow{b} \right)=\overrightarrow{c}.\overrightarrow{b} \\\ & \because \overrightarrow{a}\bot \overrightarrow{b}\Rightarrow \overrightarrow{a}.\overrightarrow{b}=0 \\\ & \therefore \overrightarrow{p}\left( \overrightarrow{r}.\overrightarrow{b} \right)=\overrightarrow{c}.\overrightarrow{b} \\\ & \Rightarrow \left( \overrightarrow{r}.\overrightarrow{b} \right)=\dfrac{\overrightarrow{c}.\overrightarrow{b}}{\overrightarrow{p}} \\\ \end{aligned}
Let us substitute the value of r→.b→\overrightarrow{r}.\overrightarrow{b} in the given equation.

& \overrightarrow{p}\overrightarrow{r}+\dfrac{\left( \overrightarrow{c}.\overrightarrow{b} \right)}{\overrightarrow{p}}\overrightarrow{a}=\overrightarrow{c} \\\ & \Rightarrow \overrightarrow{p}\overrightarrow{r}=\dfrac{\overrightarrow{p}\overrightarrow{c}-\left( \overrightarrow{c}.\overrightarrow{b} \right)\overrightarrow{a}}{\overrightarrow{p}} \\\ & \Rightarrow \overrightarrow{r}=\dfrac{\overrightarrow{p}\overrightarrow{c}-\left( \overrightarrow{c}.\overrightarrow{b} \right)\overrightarrow{a}}{{{p}^{2}}} \\\ & \Rightarrow \overrightarrow{r}=\dfrac{\overrightarrow{c}}{\overrightarrow{p}}-\dfrac{\left( \overrightarrow{c}.\overrightarrow{b} \right)\overrightarrow{a}}{{{p}^{2}}} \\\ \end{aligned}$$ Therefore, the correct option is option (A). Note: Some students get confused that for two perpendicular vectors, their cross product is zero. This is wrong and such confusion should be removed. For two perpendicular vectors, the dot product is always zero and the cross product will give a vector which is perpendicular to both the vectors.