A plane pi makes intercepts 3 and 4 respectiv | vedclass

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Question

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the intercepts on the $X, Y, Z$ axes

Vedclass · Accepted Answer

$3x + 4z = 12$

Vedclass · Answer

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the intercepts on the $X, Y, Z$ axes respectively.Given that the plane makes an intercept of $4$ on the $X$-axis $(a = 4)$ and an intercept of $3$ on the $Z$-axis $(c = 3)$.Since the plane is parallel to the $Y$-axis, it does not intersect the $Y$-axis at any finite distance, which implies the intercept $b$ on the $Y$-axis is at infinity $(b 	o \infty)$.Therefore, the term $\frac{y}{b}$ becomes $\frac{y}{\infty} = 0$.The equation of the plane becomes $\frac{x}{4} + \frac{z}{3} = 1$.Multiplying the entire equation by $12$, we get $3x + 4z = 12$.

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is:

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