The sum of intercepts of the plane 4x + 3y + | vedclass

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

Question

(A) The given equation of the plane is $4x + 3y + 2z = 2$. To find the intercepts,we convert this equation into the intercept form $\frac{x}{a} + \fra

Vedclass · Accepted Answer

$\frac{13}{6}$

Vedclass · Answer

(A) The given equation of the plane is $4x + 3y + 2z = 2$. To find the intercepts,we convert this equation into the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Dividing the entire equation by $2$,we get: $\frac{4x}{2} + \frac{3y}{2} + \frac{2z}{2} = \frac{2}{2}$ $\frac{x}{1/2} + \frac{y}{2/3} + \frac{z}{1} = 1$. Comparing this with the intercept form,the intercepts are $a = \frac{1}{2}$,$b = \frac{2}{3}$,and $c = 1$. The sum of the intercepts is $a + b + c = \frac{1}{2} + \frac{2}{3} + 1$. Taking the least common multiple $(LCM)$ of $2$ and $3$,which is $6$: $a + b + c = \frac{3 + 4 + 6}{6} = \frac{13}{6}$.

The sum of intercepts of the plane $4x + 3y + 2z = 2$ on the coordinate axes is

Similar Questions

If the angle between the planes $\bar{r} \cdot(11 \hat{i}-2 \hat{j}+\alpha \hat{k})=7$ and $\bar{r} \cdot(2 \hat{i}+4 \hat{j}-2 \hat{k})=5$ is $\frac{\pi}{2}$, then $\alpha=$

Find the Cartesian equation of the plane $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.

The angle between two planes is defined as:

The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is:

The volume (in cubic units) of the tetrahedron bounded by the plane $3x + 4y - 5z = 60$ and the three coordinate planes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module