If P - 15- -2 be a point and L - x-1-2 - y-2-1 - z-3-1 - r be a line if PQ is the distance of plane -2x - y - z -1 - 0 from P measured along a line inclined at an angle of 60o with line L and is minimum- then PQ is equal to

The correct option is A 1 unitP = (15, √(2)) be a point #160;and L = x 1/√(2) = y 2/1 = z 3/1 = r be a line #160;PQ is the distance of plane ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Plane

If P = 15, ...

Question

If P = $(15, \sqrt{2})$ be a point and L = $\frac{x - 1}{\sqrt{2}} = \frac{y - 2}{1} = \frac{z - 3}{1} = r$ be a line if PQ is the distance of plane $\sqrt{2} x + y - z - 1 = 0$ from P measured along a line inclined at an angle of $60^{o}$ with line L and is minimum, then PQ is equal to

1 unit

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2 \sqrt{3}

units

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2 units

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5 units.

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Solution

The correct option is A 1 unit
P = $(15, \sqrt{2})$ be a point
and L = $\frac{x - 1}{\sqrt{2}} = \frac{y - 2}{1} = \frac{z - 3}{1} = r$ be a line
PQ is the distance of plane $\sqrt{2} x + y - z - 1 = 0$ from P
measured along a line inclined at an angle of $60^{o}$ with line L and is minimum,
$∴$ PQ is equal to 1 units

Suggest Corrections

Similar questions

Q. Equation of lines

L_{1} : 2 x - 2 y + 3 z - 2 = 0 = x - y + z + 1 = 0

L_{2} : x + 2 y - z - 3 = 0 = 3 x - y + 2 z - 1 = 0

Distance of point

P (0, 0, 0)

from the plane containing

L_{1}

and

L_{2}

and measured along the line

x = y = z

\frac{a \sqrt{a}}{b}

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and

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Q. Let

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be the plane containing the line

L_{1} : y + z = 2,

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and is parallel to the line

L_{2} : x - z = 2, y = 0.

If the distance of the plane

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to the straight line

L : \frac{x}{1} = \frac{y}{0} = \frac{z}{- 1}

N

. Let a line be drawn from

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x + y + 2 z = 0

which meets

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and the plane

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at a point

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then

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Q. Let

P_{1} : x + y + z = 0

P_{2} : x + 2 y + 3 z = 0

P_{3} : x + 3 y + 5 z = 0

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If P = (15,√2) be a point and L = x−1√2=y−21=z−31=r be a line if PQ is the distance of plane √2x+y−z−1=0 from P measured along a line inclined at an angle of 60o with line L and is minimum, then PQ is equal to

The correct option is A 1 unitP = (15,√2) be a point and L = x−1√2=y−21=z−31=r be a line PQ is the distance of plane √2x+y−z−1=0 from P measured along a line inclined at an angle of 60o with line L and is minimum, ∴ PQ is equal to 1 units

If P = $(15, \sqrt{2})$ be a point and L = $\frac{x - 1}{\sqrt{2}} = \frac{y - 2}{1} = \frac{z - 3}{1} = r$ be a line if PQ is the distance of plane $\sqrt{2} x + y - z - 1 = 0$ from P measured along a line inclined at an angle of $60^{o}$ with line L and is minimum, then PQ is equal to