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

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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,

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

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:

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