If the vectors vec{BC} = 2hat{i} + hat{j} + h | vedclass

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

Question

(B) Let $\vec{a} = \vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$.In parallelogram $ABCD$, the side

Vedclass · Accepted Answer

$\frac{-10\sqrt{2}}{3}$

Vedclass · Answer

(B) Let $\vec{a} = \vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$.In parallelogram $ABCD$, the sides are $\vec{BC}$ and $\vec{CD}$. The diagonals are $\vec{AC} = \vec{BC} + \vec{CD} = (2\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) = 3\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{BD} = \vec{BC} - \vec{CD} = (2\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = \hat{i} - \hat{j} + 3\hat{k}$.Let $\vec{d_1} = \vec{AC} = 3\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{d_2} = \vec{BD} = \hat{i} - \hat{j} + 3\hat{k}$.$|\vec{d_1}| = \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19}$.$|\vec{d_2}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}$.$\vec{d_1} \cdot \vec{d_2} = (3)(1) + (3)(-1) + (-1)(3) = 3 - 3 - 3 = -3$.$\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-3}{\sqrt{19} \cdot \sqrt{11}} = \frac{-3}{\sqrt{209}}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{9}{209} = \frac{200}{209}$, we have $\sin 	heta = \frac{\sqrt{200}}{\sqrt{209}} = \frac{10\sqrt{2}}{\sqrt{209}}$.Therefore, $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{10\sqrt{2} / \sqrt{209}}{-3 / \sqrt{209}} = \frac{-10\sqrt{2}}{3}$.

If the vectors $\vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\vec{AC}$ and $\vec{BD}$, then $\tan \theta =$

Similar Questions

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $|p| = p, |q| = q.$ The points $R$ and $S$ divide $PQ$ internally and externally in the ratio $2 : 3$ respectively. If $\overrightarrow{OR}$ and $\overrightarrow{OS}$ are perpendicular,then:

The cosine of the angle between any two diagonals of a cube is

If $ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $,then the angle between $ \vec{a} $ and $ \vec{b} $ is: (in $^{\circ}$)

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module