p=2 hat{i}-3 hat{j}+hat{k}, q=hat{i}+hat{j}-h | vedclass

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

Question

(C) Given $p=2 \hat{i}-3 \hat{j}+\hat{k}$ and $q=\hat{i}+\hat{j}-\hat{k}$.First, calculate the dot product $p \cdot q = (2)(1) + (-3)(1) + (1)(-1) = 2

Vedclass · Accepted Answer

$\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}$

Vedclass · Answer

(C) Given $p=2 \hat{i}-3 \hat{j}+\hat{k}$ and $q=\hat{i}+\hat{j}-\hat{k}$.First, calculate the dot product $p \cdot q = (2)(1) + (-3)(1) + (1)(-1) = 2 - 3 - 1 = -2$.Calculate the magnitudes squared: $|p|^2 = 2^2 + (-3)^2 + 1^2 = 4 + 9 + 1 = 14$ and $|q|^2 = 1^2 + 1^2 + (-1)^2 = 3$.Vector $a$ (projection of $p$ on $q$) is given by $a = \frac{p \cdot q}{|q|^2} q = \frac{-2}{3}(\hat{i}+\hat{j}-\hat{k})$.Vector $b$ (projection of $q$ on $p$) is given by $b = \frac{q \cdot p}{|p|^2} p = \frac{-2}{14}(2 \hat{i}-3 \hat{j}+\hat{k}) = \frac{-1}{7}(2 \hat{i}-3 \hat{j}+\hat{k})$.Now, $a 	imes b = \left(\frac{-2}{3}ight) \left(\frac{-1}{7}ight) [(\hat{i}+\hat{j}-\hat{k}) 	imes (2 \hat{i}-3 \hat{j}+\hat{k})] = \frac{2}{21} [\hat{i}(1-3) - \hat{j}(1+2) + \hat{k}(-3-2)] = \frac{2}{21} (-2 \hat{i}-3 \hat{j}-5 \hat{k})$.Also, $a \cdot b = \left(\frac{-2}{3}ight) \left(\frac{-1}{7}ight) [(\hat{i}+\hat{j}-\hat{k}) \cdot (2 \hat{i}-3 \hat{j}+\hat{k})] = \frac{2}{21} (2 - 3 - 1) = \frac{2}{21} (-2) = \frac{-4}{21}$.Finally, $\frac{a 	imes b}{a \cdot b} = \frac{\frac{2}{21} (-2 \hat{i}-3 \hat{j}-5 \hat{k})}{\frac{-4}{21}} = \frac{-2 \hat{i}-3 \hat{j}-5 \hat{k}}{-2} = \frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}$.

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively, then $\frac{a \times b}{a \cdot b}=$

Similar Questions

If $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$, then $\theta$ is equal to

The points $A(5, -1, 1)$,$B(7, -4, 7)$,$C(1, -6, 10)$,and $D(-1, -3, 4)$ are vertices of a:

If the moduli of $a$ and $b$ are equal and the angle between them is $120^\circ$ and $a \cdot b = -8$,then $|a|$ is equal to

If $\overline{p}=2 \hat{i}+\hat{k}$,$\overline{q}=\hat{i}+\hat{j}+\hat{k}$,$\overline{r}=4 \hat{i}-3 \hat{j}+7 \hat{k}$ and a vector $\overline{m}$ is such that $\overline{m} \times \overline{q}=\overline{r} \times \overline{q}$ and $\overline{m} \cdot \overline{p}=0$,then $\overline{m} = \dots$

If $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal,then the value of $\lambda$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module