Question

If vector A vector = i cap + c j cap + 5 k cap and vector B = 2 i cap + j cap - k cap are perpendicular, then calculate the value of c.

Answer 1

A=Icap+ cjcap+5kcap
B= 2icap +jcap-kcap
As both are perenducular the dot protect is zero a.b= 0
(Icap+cjcap +5kcap).(2icap+jcap-kcap)
= 0
2+c-5 = 0
C= 3

Answer 2

The value of c is 3.

Given data suggest that the $\vec{A} \text { and } \overrightarrow{\mathrm{B}}$ are two vector which are perpendicular to each other. The $\overrightarrow{A}=\hat{\imath}+c \widehat{\jmath}+5 \hat{k} \text { and } \overrightarrow{\mathrm{B}}=2 \hat{\imath}+\mathrm{c} \widehat{\jmath}-\hat{k}$.

These both are perpendicular to each other which implies that their dot product will be zero.  

$\Rightarrow \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0$.

$\Rightarrow(\hat{\imath}+c(j)+5 \hat{k}) \cdot(2 \hat{\imath}+c(\hat{\jmath})-\hat{k})=0$

$\Rightarrow(i \times 2 i+c \times 1+[5 \times-1])=0$

$\begin{array}{l}{\Rightarrow 2+c-5=0} \\ {\Rightarrow c=5-2} \\ \bold{{\Rightarrow c=3}\end{array}}$.