Let overline{a}=2 hat{i}+hat{j}-2 hat{k} and | vedclass

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

Question

(A) First,calculate the cross product $\overline{a} 	imes \overline{b}$:$\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) First,calculate the cross product $\overline{a} 	imes \overline{b}$:$\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.The magnitude is $|\overline{a} 	imes \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$,squaring both sides gives $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{c} \cdot \overline{a}) = 8$.Since $|\overline{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$ and $\overline{a} \cdot \overline{c} = |\overline{c}|$,we have $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$.$|\overline{c}|^2 - 2|\overline{c}| + 1 = 0 \Rightarrow (|\overline{c}| - 1)^2 = 0 \Rightarrow |\overline{c}| = 1$.The magnitude of the cross product is $|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(30^{\circ})$.Substituting the values: $3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

Similar Questions

The unit vector perpendicular to the vector $\hat{i}-2 \hat{j}+3 \hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}-\hat{k}$ is

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

The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ is

If the area of the triangle with vertices $\hat{i}+y \hat{j}$,$\hat{i}+2 \hat{k}$,and $3 \hat{j}+\hat{k}$ is $\sqrt{6}$ sq. units,then the values of $y$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module