A plane meets the coordinate axes at P, Q, an | vedclass

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

Question

(A) Let the intercepts of the plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.The coordinates of the vertices are $P(a, 0, 0), Q(0,

Vedclass · Accepted Answer

$r \cdot (20i - 8j + 5k) = 120$

Vedclass · Answer

(A) Let the intercepts of the plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.The coordinates of the vertices are $P(a, 0, 0), Q(0, b, 0),$ and $R(0, 0, c)$.The centroid of $\Delta PQR$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.Given the centroid is $2i - 5j + 8k$,we have $\frac{a}{3} = 2, \frac{b}{3} = -5, \frac{c}{3} = 8$.Thus,$a = 6, b = -15, c = 24$.The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values,we get $\frac{x}{6} + \frac{y}{-15} + \frac{z}{24} = 1$.Multiplying by the least common multiple of $6, 15,$ and $24$,which is $120$,we get $20x - 8y + 5z = 120$.In vector form,this is $r \cdot (20i - 8j + 5k) = 120$.

$A$ plane meets the coordinate axes at $P, Q,$ and $R$ such that the position vector of the centroid of $\Delta PQR$ is $2i - 5j + 8k$. Then the equation of the plane is:

Similar Questions

If the foot of the perpendicular from $(0,0,0)$ to the plane is $(1,2,2)$, then the equation of the plane is

The vector equation of the plane passing through the point $2\hat{i} - \hat{j} - 4\hat{k}$ and parallel to the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) - 7 = 0$ is:

Find the Cartesian equation of the plane passing through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$.

$P$ and $Q$ are points on the straight line passing through the point $A(3 \hat{i}+\hat{j}-\hat{k})$ and parallel to the vector $\vec{v} = 2 \hat{i}-\hat{j}+2 \hat{k}$. If $AP = AQ = 3$,then the vector equation of the plane $OPQ$ is:

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module