Let overline{a}=3 hat{i}-alpha hat{j}+hat{k} | vedclass

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

Question

(B) The area of a parallelogram with adjacent sides $\overline{a}$ and $\overline{b}$ is given by $|\overline{a} 	imes \overline{b}|$. First,we calcu

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The area of a parallelogram with adjacent sides $\overline{a}$ and $\overline{b}$ is given by $|\overline{a} 	imes \overline{b}|$. First,we calculate the cross product $\overline{a} 	imes \overline{b}$: $\overline{a} 	imes \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -\alpha & 1 \ 1 & \alpha & 3 \end{vmatrix} = \hat{i}(-3\alpha - \alpha) - \hat{j}(9 - 1) + \hat{k}(3\alpha + \alpha) = -4\alpha \hat{i} - 8 \hat{j} + 4\alpha \hat{k}$. The magnitude is $|\overline{a} 	imes \overline{b}| = \sqrt{(-4\alpha)^2 + (-8)^2 + (4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$. Given the area is $8\sqrt{3}$,we have $\sqrt{32\alpha^2 + 64} = 8\sqrt{3}$. Squaring both sides: $32\alpha^2 + 64 = 64 	imes 3 = 192$. $32\alpha^2 = 128 \Rightarrow \alpha^2 = 4$. Now,calculate the dot product $\overline{a} \cdot \overline{b} = (3\hat{i} - \alpha\hat{j} + \hat{k}) \cdot (\hat{i} + \alpha\hat{j} + 3\hat{k}) = 3(1) - \alpha(\alpha) + 1(3) = 3 - \alpha^2 + 3 = 6 - \alpha^2$. Substituting $\alpha^2 = 4$,we get $\overline{a} \cdot \overline{b} = 6 - 4 = 2$.

Let $\overline{a}=3 \hat{i}-\alpha \hat{j}+\hat{k}$ and $\overline{b}=\hat{i}+\alpha \hat{j}+3 \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overline{a}$ and $\overline{b}$ is $8 \sqrt{3}$ sq. units,then $\overline{a} \cdot \overline{b}$ is equal to

Similar Questions

If $a$,$b$,$c$ are the $p^{th}$,$q^{th}$,$r^{th}$ terms of an $A.P.$ and $\vec x = (q - r)\hat i + (r - p)\hat j + (p - q)\hat k$ and $\vec y = a\hat i + b\hat j + c\hat k$,then:

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by:

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module