If the angle between the planes x-2y+3z-5=0 a | vedclass

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

Question

(B) The normal vectors to the planes are $\vec{n_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{n_2} = \hat{i} + \alpha\hat{j} + 2\hat{k}$.The angle $\

Vedclass · Accepted Answer

$\frac{62}{55}$

Vedclass · Answer

(B) The normal vectors to the planes are $\vec{n_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{n_2} = \hat{i} + \alpha\hat{j} + 2\hat{k}$.The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.Given $\cos 	heta = \frac{1}{14}$,we have $\frac{|(1)(1) + (-2)(\alpha) + (3)(2)|}{\sqrt{1^2 + (-2)^2 + 3^2} \sqrt{1^2 + \alpha^2 + 2^2}} = \frac{1}{14}$.$\frac{|7 - 2\alpha|}{\sqrt{14} \sqrt{\alpha^2 + 5}} = \frac{1}{14}$.Squaring both sides: $\frac{(7 - 2\alpha)^2}{14(\alpha^2 + 5)} = \frac{1}{196}$.$\frac{49 - 28\alpha + 4\alpha^2}{\alpha^2 + 5} = \frac{14}{196} = \frac{1}{14}$.$14(4\alpha^2 - 28\alpha + 49) = \alpha^2 + 5$.$56\alpha^2 - 392\alpha + 686 = \alpha^2 + 5$.$55\alpha^2 - 392\alpha + 681 = 0$.Let the roots be $\alpha_1$ and $\alpha_2$. The difference is $|\alpha_1 - \alpha_2| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{(-392)^2 - 4(55)(681)}}{55}$.$D = 153664 - 149820 = 3844$.$\sqrt{D} = \sqrt{3844} = 62$.Difference $= \frac{62}{55}$.

If the angle between the planes $x-2y+3z-5=0$ and $x+\alpha y+2z+7=0$ is $\cos^{-1}\left(\frac{1}{14}\right)$,then the difference between the values of $\alpha$ is

Similar Questions

The equation of the plane passing through the points $(2,1,0)$,$(3,2,-2)$,and $(3,1,7)$ is

The equation of the plane which contains the $y$-axis and passes through the point $(1, 2, 3)$ is:

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$,$2\hat{i}+3\hat{j}-4\hat{k}$,and $3\hat{i}-4\hat{j}+5\hat{k}$ is

$A$ plane passing through the points $A(2, 3, 5)$ and $B(-3, -5, -7)$ is perpendicular to the plane $x - y + z = 1$. Which of the following points lies on this plane?

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module