Foot of perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 is: A (5, -1, 4) B (2, -3, 4) C (7, -1, 3) D (5, -2, 3)
Question
Foot of perpendicular drawn from the origin to the plane $2x - 3y + 4z = 29$ is:
- A $(5, -1, 4)$
- B $(2, -3, 4)$
- C $(7, -1, 3)$
- D $(5, -2, 3)$
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Answer
:
The correct option is B.
We start by recognizing that the problem asks for the foot of the perpendicular drawn from the origin to the plane given by the equation:
$$ 2x - 3y + 4z = 29 $$
Assume the coordinates of the foot of the perpendicular from the origin to the plane are $(a, b, c)$. Since this point lies on the plane and the direction ratios of the line perpendicular to the plane are proportional to the coefficients of the plane equation, we can say that:
$$ \frac{a}{2} = \frac{b}{-3} = \frac{c}{4} $$
Also, this point must satisfy the given plane equation:
$$ 2a - 3b + 4c = 29 $$
We use the equal proportional ratios set to find compatible $a$, $b$, and $c$. By setting $\frac{a}{2} = \frac{b}{-3} = \frac{c}{4} = k$, we get:
$$ a = 2k, \quad b = -3k, \quad c = 4k $$
Substitute these into the plane equation:
$$ 2(2k) - 3(-3k) + 4(4k) = 29 $$
Solving for $k$:
$$ 4k + 9k + 16k = 29 \ 29k = 29 \ k = 1 $$
Thus, substituting $k = 1$ back into the proportional relationships, we find:
$$ a = 2(1) = 2, \quad b = -3(1) = -3, \quad c = 4(1) = 4 $$
Therefore, the coordinates of the foot of the perpendicular are:
$$ (2, -3, 4) $$
Hence, the correct answer is:
- B $(2, -3, 4)$
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