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Centre of Ellipse

If the line, ...

Question

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point P, then the distance of P from the origin is:

\frac{7}{2}

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\frac{9}{2}

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2 \sqrt{5}

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\frac{\sqrt{5}}{2}

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Solution

The correct option is B $\frac{9}{2}$
$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4} = k [L e t] \Rightarrow x = 2 k + 1, y = 3 k - 1, z = 4 k + 2$
Given point satisfies the equation of plane $x + 2 y + 3 z = 15$
$\Rightarrow 2 k + 1 + 2 (3 k - 1) + 3 (4 k + 2) = 15 \Rightarrow 20 k = 10 \Rightarrow k = \frac{1}{2} ∴ P o i n t P i s (2, \frac{1}{2}, 4) .$
Distance from Origin $= \sqrt{4 + \frac{1}{4} + 16} = \sqrt{\frac{81}{4}} = \frac{9}{2}$

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Similar questions

Q. If the line

\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}

meets the plane

x + 2 y + 3 z = 14

P

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The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point (3,1,-1) is

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Q. Two lines whose equation are

L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{λ}

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If the line, x−12=y+13=z−24 meets the plane, x+2y+3z=15 at a point P, then the distance of P from the origin is:

The correct option is B 92x−12=y+13=z−24=k [Let]⇒x=2k+1,y=3k−1,z=4k+2 Given point satisfies the equation of plane x+2y+3z=15 ⇒2k+1+2(3k−1)+3(4k+2)=15⇒20k=10⇒k=12∴Point P is (2,12,4). Distance from Origin=√4+14+16=√814=92

If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point P, then the distance of P from the origin is: