For some real number lambda,if the area of th | vedclass

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

Question

(C) Given $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$.Area of the triangle $= \frac{1}{2} |\vec{a} 	i

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$.Area of the triangle $= \frac{1}{2} |\vec{a} 	imes \vec{b}| = \frac{\sqrt{195}}{2}$.Thus,$|\vec{a} 	imes \vec{b}|^2 = 195$ ...$(i)$.Now,$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & \lambda \ \lambda & 1 & -3 \end{vmatrix} = \hat{i}(3-\lambda) - \hat{j}(-9-\lambda^2) + \hat{k}(3+\lambda) = (3-\lambda)\hat{i} + (9+\lambda^2)\hat{j} + (3+\lambda)\hat{k}$.$|\vec{a} 	imes \vec{b}|^2 = (3-\lambda)^2 + (9+\lambda^2)^2 + (3+\lambda)^2 = 195$.Expanding the terms: $(9 + \lambda^2 - 6\lambda) + (81 + \lambda^4 + 18\lambda^2) + (9 + \lambda^2 + 6\lambda) = 195$.$\lambda^4 + 20\lambda^2 + 99 = 195 \Rightarrow \lambda^4 + 20\lambda^2 - 96 = 0$.Let $t = \lambda^2$,then $t^2 + 20t - 96 = 0 \Rightarrow (t+24)(t-4) = 0$.Since $\lambda$ is a real number,$\lambda^2 = 4$ (as $\lambda^2 = -24$ is impossible).Thus,$\lambda = \pm 2$. The number of distinct values of $\lambda$ is $2$.

For some real number $\lambda$,if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$,then the number of distinct possible values of $\lambda$ is

Similar Questions

The area of a parallelogram whose adjacent sides are $i - 2j + 3k$ and $2i + j - 4k$ is:

$\vec{a}, \vec{b}, \vec{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\vec{a}, \vec{b})=(\vec{b}, \vec{c})=(\vec{c}, \vec{a})=\frac{\pi}{3}$. If $\vec{x}=\vec{a} \times(\vec{b} \times \vec{c})$ and $\vec{y}=\vec{b} \times(\vec{c} \times \vec{a})$, then

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2, \bar{b} \cdot \bar{c}=8$ and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then the value of $|\bar{a} \times(\bar{b} \times \bar{c})|$ is

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b},$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}.$

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$, then $\lambda+\mu$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module