A vector overrightarrow{V} in the first octan | vedclass

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

Question

(C) Let the unit vector be $\hat{v} = \cos 60^{\circ} \hat{i} + \cos 45^{\circ} \hat{j} + \cos \gamma \hat{k}$.Since $\cos^2 \alpha + \cos^2 \beta + \

Vedclass · Accepted Answer

$a + \sqrt{2} b + c = 1$

Vedclass · Answer

(C) Let the unit vector be $\hat{v} = \cos 60^{\circ} \hat{i} + \cos 45^{\circ} \hat{j} + \cos \gamma \hat{k}$.Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,we have $\cos^2 60^{\circ} + \cos^2 45^{\circ} + \cos^2 \gamma = 1$.$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.$\Rightarrow \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.Since $\gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$.Thus,the normal vector to the plane is $\vec{n} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$.The equation of the plane passing through $(\sqrt{2}, -1, 1)$ is $\frac{1}{2}(x - \sqrt{2}) + \frac{1}{\sqrt{2}}(y + 1) + \frac{1}{2}(z - 1) = 0$.Multiplying by $2$,we get $(x - \sqrt{2}) + \sqrt{2}(y + 1) + (z - 1) = 0$.$\Rightarrow x - \sqrt{2} + \sqrt{2} y + \sqrt{2} + z - 1 = 0$.$\Rightarrow x + \sqrt{2} y + z = 1$.Since the point $(a, b, c)$ lies on the plane,we have $a + \sqrt{2} b + c = 1$.

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then:

Similar Questions

If the plane $3x - 2y - z - 18 = 0$ meets the coordinate axes at $A, B, C$,then the centroid of $\triangle ABC$ is

$A$ point on the plane passing through the points $(\sqrt{2}, 1, 4)$,$(0, -1, 0)$ and $(0, 0, 1)$ is

The distance of the plane $2x - y - 2z - 9 = 0$ from the origin is $d$ units.

The equation of the plane whose intercepts on $x, y, z$ axes are $1, 2, 4$ respectively is

The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module