The plane frac{x}{2} + frac{y}{3} + frac{z}{4 | vedclass

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

Question

(C) The plane equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Setting $y=0, z=0$ gives $x=2$, so $A = (2, 0, 0)$. Setting $x=0, z=0$ gives

Vedclass · Accepted Answer

$\sqrt{61}$

Vedclass · Answer

(C) The plane equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Setting $y=0, z=0$ gives $x=2$, so $A = (2, 0, 0)$. Setting $x=0, z=0$ gives $y=3$, so $B = (0, 3, 0)$. Setting $x=0, y=0$ gives $z=4$, so $C = (0, 0, 4)$. The vectors $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 4)$. The cross product $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12 - 0) - \hat{j}(-8 - 0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$. The area of triangle $ABC$ is $\frac{1}{2} |\vec{AB} 	imes \vec{AC}| = \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 	imes 61} = \sqrt{61}$.

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes at the points $A, B, C$. Find the area of triangle $ABC$.

Similar Questions

If a plane meets the coordinate axes at $A, B$,and $C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$,then the equation of the plane is:

The length of the perpendicular from the origin to the plane $\bar{r} \cdot (3 \hat{i} - 4 \hat{j} + 12 \hat{k}) = 8$ is

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

Let two planes be $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Find the equation of the angle bisector plane of $P_1$ and $P_2$ that does not contain the origin.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module