If the plane frac{x}{2} + frac{y}{3} + frac{z | vedclass

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

Question

(D) The given plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$. The intercepts on the axes are $a=2, b=3, c=3$. Thus,the coordinates of the vert

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The given plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$. The intercepts on the axes are $a=2, b=3, c=3$. Thus,the coordinates of the vertices are $A(2, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 3)$. The vectors forming the sides are $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 3)$. The area of $\Delta ABC$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. Calculating the cross product: $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 3 \end{vmatrix} = \hat{i}(9 - 0) - \hat{j}(-6 - 0) + \hat{k}(0 - (-6)) = 9\hat{i} + 6\hat{j} + 6\hat{k}$. The magnitude is $|\vec{AB} 	imes \vec{AC}| = \sqrt{9^2 + 6^2 + 6^2} = \sqrt{81 + 36 + 36} = \sqrt{153} = 3\sqrt{17}$. Therefore,the area is $\frac{1}{2} 	imes 3\sqrt{17} = \frac{3\sqrt{17}}{2}$ sq. units. Since this value is not among the options,the correct choice is $D$.

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

Similar Questions

The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is

Find the equation of the plane with intercepts $2, 3$ and $4$ on the $x, y$ and $z$-axis respectively.

The angle between the planes $x+y+2z=6$ and $2x-y+z=9$ is:

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them: $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module