If overline{a}, overline{b}, overline{c} are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors,we have $\overline{a} \cdot \overline{b} = 0, \overline{b

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors,we have $\overline{a} \cdot \overline{b} = 0, \overline{b} \cdot \overline{c} = 0, \overline{c} \cdot \overline{a} = 0$ and magnitudes $|\overline{a}|=1, |\overline{b}|=2, |\overline{c}|=3$. The scalar triple product is defined as $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}] = (\overline{a}+\overline{b}+\overline{c}) \cdot ((\overline{b}-\overline{a}) 	imes \overline{c})$. Expanding the cross product: $(\overline{b}-\overline{a}) 	imes \overline{c} = \overline{b} 	imes \overline{c} - \overline{a} 	imes \overline{c}$. Now,substitute this back: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{b} 	imes \overline{c} - \overline{a} 	imes \overline{c})$. Since $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular,$\overline{a} \cdot (\overline{b} 	imes \overline{c}) = |\overline{a}| |\overline{b}| |\overline{c}| = 1 	imes 2 	imes 3 = 6$. Also,$\overline{b} \cdot (\overline{b} 	imes \overline{c}) = 0$ and $\overline{c} \cdot (\overline{b} 	imes \overline{c}) = 0$. Similarly,$\overline{a} \cdot (\overline{a} 	imes \overline{c}) = 0, \overline{b} \cdot (\overline{a} 	imes \overline{c}) = 0, \overline{c} \cdot (\overline{a} 	imes \overline{c}) = 0$. The expression simplifies to: $\overline{a} \cdot (\overline{b} 	imes \overline{c}) - \overline{a} \cdot (\overline{a} 	imes \overline{c}) + \overline{b} \cdot (\overline{b} 	imes \overline{c}) - \overline{b} \cdot (\overline{a} 	imes \overline{c}) + \overline{c} \cdot (\overline{b} 	imes \overline{c}) - \overline{c} \cdot (\overline{a} 	imes \overline{c})$. This reduces to $\overline{a} \cdot (\overline{b} 	imes \overline{c}) - (-\overline{b} \cdot (\overline{a} 	imes \overline{c})) = [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] = [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = 2[\overline{a} \overline{b} \overline{c}]$. Since $[\overline{a} \overline{b} \overline{c}] = 1 	imes 2 	imes 3 = 6$,the final value is $2 	imes 6 = 12$.

If $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then the value of $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}]$ is

Similar Questions

If the origin and the points $(1, 2, 3)$,$(2, 3, 4)$,and $(x, y, z)$ are coplanar,then

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$,$\hat{i}+\hat{j}+\hat{k}$,and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

If the scalar triple product of the vectors $-3 \hat{i}+7 \hat{j}-3 \hat{k}$,$3 \hat{i}-7 \hat{j}+\lambda \hat{k}$ and $7 \hat{i}-5 \hat{j}-3 \hat{k}$ is $272$,then $\lambda = \ldots$

The value of $m$,if the vectors $\hat{\imath}-\hat{\jmath}-6 \hat{k}$,$\hat{\imath}-3 \hat{\jmath}+4 \hat{k}$,and $2 \hat{\imath}-5 \hat{\jmath}+m \hat{k}$ are coplanar,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module