If bar{u}, bar{v}, and bar{w} are three non-c | vedclass

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

Question

(C) Let the expression be $E = (\bar{u} + \bar{v} - \bar{w}) \cdot ((\bar{u} - \bar{v}) 	imes (\bar{v} - \bar{w}))$.First,compute the cross product:

Vedclass · Accepted Answer

$\bar{u} \cdot (\bar{v} 	imes \bar{w})$

Vedclass · Answer

(C) Let the expression be $E = (\bar{u} + \bar{v} - \bar{w}) \cdot ((\bar{u} - \bar{v}) 	imes (\bar{v} - \bar{w}))$.First,compute the cross product: $(\bar{u} - \bar{v}) 	imes (\bar{v} - \bar{w}) = \bar{u} 	imes \bar{v} - \bar{u} 	imes \bar{w} - \bar{v} 	imes \bar{v} + \bar{v} 	imes \bar{w}$.Since $\bar{v} 	imes \bar{v} = 0$,this simplifies to $\bar{u} 	imes \bar{v} + \bar{w} 	imes \bar{u} + \bar{v} 	imes \bar{w}$.Now,take the dot product with $(\bar{u} + \bar{v} - \bar{w})$:$E = (\bar{u} + \bar{v} - \bar{w}) \cdot (\bar{u} 	imes \bar{v} + \bar{w} 	imes \bar{u} + \bar{v} 	imes \bar{w})$.Distributing the dot product:$= \bar{u} \cdot (\bar{u} 	imes \bar{v}) + \bar{u} \cdot (\bar{w} 	imes \bar{u}) + \bar{u} \cdot (\bar{v} 	imes \bar{w}) + \bar{v} \cdot (\bar{u} 	imes \bar{v}) + \bar{v} \cdot (\bar{w} 	imes \bar{u}) + \bar{v} \cdot (\bar{v} 	imes \bar{w}) - \bar{w} \cdot (\bar{u} 	imes \bar{v}) - \bar{w} \cdot (\bar{w} 	imes \bar{u}) - \bar{w} \cdot (\bar{v} 	imes \bar{w})$.Using the property that scalar triple product is zero if any two vectors are the same:$= 0 + 0 + [\bar{u} \bar{v} \bar{w}] + 0 + [\bar{v} \bar{w} \bar{u}] + 0 - [\bar{w} \bar{u} \bar{v}] - 0 - 0$.Since $[\bar{u} \bar{v} \bar{w}] = [\bar{v} \bar{w} \bar{u}] = [\bar{w} \bar{u} \bar{v}]$,we get:$E = [\bar{u} \bar{v} \bar{w}] + [\bar{u} \bar{v} \bar{w}] - [\bar{u} \bar{v} \bar{w}] = [\bar{u} \bar{v} \bar{w}] = \bar{u} \cdot (\bar{v} 	imes \bar{w})$.

If $\bar{u}, \bar{v},$ and $\bar{w}$ are three non-coplanar vectors,then $(\bar{u} + \bar{v} - \bar{w}) \cdot (\bar{u} - \bar{v}) \times (\bar{v} - \bar{w}) = \dots$

Similar Questions

Let $\overline{a} = \hat{i} + \hat{j} + \hat{k}$,$\overline{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overline{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$. If the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,then $x = \dots$

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$,$\hat{i}+\hat{k}$,and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda$ is equal to

The points $A(4, 5, 1)$,$B(0, -1, -1)$,$C(3, 9, 4)$,and $D(-4, 4, 4)$ are:

If $a, b, c$ are non-coplanar vectors and $\lambda$ is a real number,then the vectors $a + 2b + 3c, \lambda b + 4c$ and $(2\lambda - 1)c$ are non-coplanar for

The value of $(\vec{a} + 2\vec{b} - \vec{c}) \cdot \{(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module