The coordinates of the foot of the perpendicu | vedclass

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(C) Given,the equation of the plane is $2x - 3y + 4z = 29$.Since $OP$ is perpendicular to the plane,the direction ratios of $OP$ are the same as the n

Vedclass · Accepted Answer

$(2, -3, 4)$

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(C) Given,the equation of the plane is $2x - 3y + 4z = 29$.Since $OP$ is perpendicular to the plane,the direction ratios of $OP$ are the same as the normal vector to the plane,which is $\langle 2, -3, 4 \rangle$.The equation of the line $OP$ passing through the origin $(0, 0, 0)$ with direction ratios $\langle 2, -3, 4 \rangle$ is given by:$\frac{x - 0}{2} = \frac{y - 0}{-3} = \frac{z - 0}{4} = \lambda$Thus,the general coordinates of any point $P$ on the line are $(2\lambda, -3\lambda, 4\lambda)$.Since point $P$ lies on the plane $2x - 3y + 4z = 29$,we substitute these coordinates into the plane equation:$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) = 29$$4\lambda + 9\lambda + 16\lambda = 29$$29\lambda = 29$$\lambda = 1$Substituting $\lambda = 1$ back into the coordinates of $P$,we get:$P = (2(1), -3(1), 4(1)) = (2, -3, 4)$.Therefore,the coordinates of the foot of the perpendicular are $(2, -3, 4)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z = 29$ are

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