Evaluate: vec{a} cdot {(vec{b} + vec{c}) time | vedclass

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

Question

(A) We need to evaluate the expression $\vec{a} \cdot \{(\vec{b} + \vec{c}) 	imes (\vec{a} + \vec{b} + \vec{c})\}$.Using the distributive property of

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We need to evaluate the expression $\vec{a} \cdot \{(\vec{b} + \vec{c}) 	imes (\vec{a} + \vec{b} + \vec{c})\}$.Using the distributive property of the cross product:$(\vec{b} + \vec{c}) 	imes (\vec{a} + \vec{b} + \vec{c}) = \vec{b} 	imes \vec{a} + \vec{b} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a} + \vec{c} 	imes \vec{b} + \vec{c} 	imes \vec{c}$.Since $\vec{x} 	imes \vec{x} = 0$,this simplifies to:$= \vec{b} 	imes \vec{a} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a} + \vec{c} 	imes \vec{b}$.Now,taking the dot product with $\vec{a}$:$= \vec{a} \cdot (\vec{b} 	imes \vec{a}) + \vec{a} \cdot (\vec{b} 	imes \vec{c}) + \vec{a} \cdot (\vec{c} 	imes \vec{a}) + \vec{a} \cdot (\vec{c} 	imes \vec{b})$.Using the definition of the scalar triple product $[\vec{x} \vec{y} \vec{z}] = \vec{x} \cdot (\vec{y} 	imes \vec{z})$:$= [\vec{a} \vec{b} \vec{a}] + [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{c} \vec{a}] + [\vec{a} \vec{c} \vec{b}]$.Since any scalar triple product with two identical vectors is $0$:$= 0 + [\vec{a} \vec{b} \vec{c}] + 0 + [\vec{a} \vec{c} \vec{b}]$.Using the property $[\vec{a} \vec{c} \vec{b}] = -[\vec{a} \vec{b} \vec{c}]$:$= [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0$.

Evaluate: $\vec{a} \cdot \{(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})\}$

Similar Questions

Let $OA, OB, OC$ be the co-terminal edges of a rectangular parallelopiped of volume $V$ and let $P$ be the vertex opposite to $O$. Then, $[\overrightarrow{AP} \overrightarrow{BP} \overrightarrow{CP}]$ is equal to

If the points having the position vectors $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}+3 \hat{j}-4 \hat{k}, -\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

If $\overline{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$,$\overline{b}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ and $\overline{c}=3 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k}$ are coplanar,then $\lambda$ is the root of the equation

If $\bar{a}=\hat{i}+5 \hat{k}, \bar{b}=2 \hat{i}+3 \hat{k}, \bar{c}=4 \hat{i}-\hat{j}+2 \hat{k}$ and $\bar{d}=\hat{i}-\hat{j}$,then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$

Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

Vedclass Test Series

Exam Paper Generator

Online Exam Module