Let overrightarrow{a}=2 hat{i}-3 hat{j}+4 hat | vedclass

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

Question

(B) Given $\vec{a} 	imes (\vec{b} + \vec{c}) + \vec{b} 	imes \vec{c} = \hat{i} + 8 \hat{j} + 13 \hat{k}$.Expanding this,we get $\vec{a} 	imes \vec{

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(B) Given $\vec{a} 	imes (\vec{b} + \vec{c}) + \vec{b} 	imes \vec{c} = \hat{i} + 8 \hat{j} + 13 \hat{k}$.Expanding this,we get $\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} + \vec{b} 	imes \vec{c} = \hat{i} + 8 \hat{j} + 13 \hat{k}$.Taking the cross product with $\vec{a}$ on both sides:$\vec{a} 	imes (\vec{a} 	imes \vec{b}) + \vec{a} 	imes (\vec{a} 	imes \vec{c}) + \vec{a} 	imes (\vec{b} 	imes \vec{c}) = \vec{a} 	imes (\hat{i} + 8 \hat{j} + 13 \hat{k})$.Using the vector triple product identity $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$:$(\vec{a} \cdot \vec{b}) \vec{a} - |\vec{a}|^2 \vec{b} + (\vec{a} \cdot \vec{c}) \vec{a} - |\vec{a}|^2 \vec{c} + (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{a} 	imes (\hat{i} + 8 \hat{j} + 13 \hat{k})$.Given $\vec{a} = (2, -3, 4)$,$\vec{b} = (3, 4, -5)$,$|\vec{a}|^2 = 4+9+16 = 29$,$|\vec{b}|^2 = 9+16+25 = 50$,$\vec{a} \cdot \vec{b} = 6-12-20 = -26$,and $\vec{a} \cdot \vec{c} = 13$.Substituting these values:$-26 \vec{a} - 29 \vec{b} + 13 \vec{a} - 29 \vec{c} + 13 \vec{b} - (-26) \vec{c} = \vec{a} 	imes (\hat{i} + 8 \hat{j} + 13 \hat{k})$.$-13 \vec{a} - 16 \vec{b} - 3 \vec{c} = \vec{a} 	imes (\hat{i} + 8 \hat{j} + 13 \hat{k})$.Taking the dot product with $\vec{b}$ on both sides:$-13 (\vec{a} \cdot \vec{b}) - 16 |\vec{b}|^2 - 3 (\vec{b} \cdot \vec{c}) = [\vec{a}, \hat{i} + 8 \hat{j} + 13 \hat{k}, \vec{b}]$.$-13 (-26) - 16 (50) - 3 (\vec{b} \cdot \vec{c}) = \begin{vmatrix} 2 & -3 & 4 \ 1 & 8 & 13 \ 3 & 4 & -5 \end{vmatrix}$.$338 - 800 - 3 (\vec{b} \cdot \vec{c}) = 2(-40-52) + 3(-5-39) + 4(4-24) = 2(-92) + 3(-44) + 4(-20) = -184 - 132 - 80 = -396$.$-462 - 3 (\vec{b} \cdot \vec{c}) = -396 \Rightarrow -3 (\vec{b} \cdot \vec{c}) = 66 \Rightarrow \vec{b} \cdot \vec{c} = -22$.Therefore,$24 - (\vec{b} \cdot \vec{c}) = 24 - (-22) = 46$.

Similar Questions

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$,$\vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to

For any three vectors $a, b, c$,the condition $a \times (b \times c) = (a \times b) \times c$ holds if:

Let $\vec{a}=-5 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}$ and $\vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\bar{b} + \bar{c}}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is:

If $a=(1,2,3), b=(2,-1,1), c=(3,2,1)$ and $a \times(b \times c)=\alpha a+\beta b+\gamma c$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module