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

(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$. Squaring both sides,we get $|\overline{c}|^2+|\overline{a}|^2-2(\overline{a} \cdot \overline{c})=8$. Since $|\overline{a}|^2 = 2^2+1^2+(-2)^2 = 9$ and $\overline{a} \cdot \overline{c}=|\overline{c}|$,the equation becomes $|\overline{c}|^2+9-2|\overline{c}|=8$. This simplifies to $(|\overline{c}|-1)^2=0$,so $|\overline{c}|=1$. Next,calculate $\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i}-2\hat{j}+\hat{k}$. The magnitude is $|\overline{a} 	imes \overline{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{4+4+1} = 3$. Finally,$|(\overline{a} 	imes \overline{b}) 	imes \overline{c}| = |\overline{a} 	imes \overline{b}| |\overline{c}| \sin(30^{\circ}) = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.

Similar Questions

If the area of the parallelogram with $a$ and $b$ as two adjacent sides is $15$ sq units,then the area of the parallelogram having $3a+2b$ and $a+3b$ as two adjacent sides in sq units is

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$, then $x=$

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines,then the direction cosines of the line which is perpendicular to both the lines will be:

The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}$ is

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. If $\vec c$ is a vector such that $\vec a \cdot \vec c + 2|\vec c| = 0$ and $|\vec c - \vec a| = \sqrt{14}$,and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$,then $|(\vec a \times \vec b) \times \vec c|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module