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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Let P be th...

Question

Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the xy-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to:

63 \sqrt{5}

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205 \sqrt{5}

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17 / \sqrt{5}

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11 / \sqrt{5}

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Solution

The correct option is A $11 / \sqrt{5}$
$λ (x + y + z - 6) + 2 x + 3 y + z + 5 = 0$
$(λ + 2) x + (λ + 3) y + (λ + 1) z + 5 - 6 λ = 0 λ + 1 = 0 \Rightarrow λ = - 1$
$P : x + 2 y + 11 = 0$
perpendicular distance $= \frac{11}{\sqrt{5}}$

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Similar questions

Q. Let

P

be a plane, which contains the line of intersection of the planes,

x + y + z - 6 = 0

and

2 x + 3 y + z + 5 = 0

and it is perpendicular to the

x y

-plane. Then the distance of the point

(0, 0, 256)

from

P

is equal to :

Q. The equation of the plane which contains the line of intersection of the planes

x + y + z - 6 = 0

and

2 x + 3 y + z + 5 = 0

and perpendicular to the

x y

plane is:

Q. Let plane

P = 0

pass through the intersection of planes

2 x - y + z - 3 = 0

and

3 x + y + z - 5 = 0

. If the distance of plane

P = 0

from

(2, 1, - 1)

\frac{1}{\sqrt{6}}

then its actuation can be:

Q. Find the vector equation of the through the line of intersection of the planes

x + y + z = 1

and

2 x + 3 y + 4 z = 5

which is perpendicular to plane

x - y + z = 0

. Hence find whether the plane thus contains the line

\frac{x - 2}{5} = \frac{y - 3}{4} = \frac{z}{5}

or not.

Find the equation of the plane which is perpendicular to the plane 5x+3y+6z+8=0 and which contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0.

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Let P be the plane, which contains the line of intersection of the planes, x+y+z−6=0 and 2x+3y+z+5=0 and it is perpendicular to the xy-plane. Then the distance of the point (0,0,256) from P is equal to:

The correct option is A 11/√5λ(x+y+z−6)+2x+3y+z+5=0(λ+2)x+(λ+3)y+(λ+1)z+5−6λ=0λ+1=0⇒λ=−1P:x+2y+11=0perpendicular distance =11√5

Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the xy-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to:

The correct option is A $11 / \sqrt{5}$
$λ (x + y + z - 6) + 2 x + 3 y + z + 5 = 0$
$(λ + 2) x + (λ + 3) y + (λ + 1) z + 5 - 6 λ = 0 λ + 1 = 0 \Rightarrow λ = - 1$
$P : x + 2 y + 11 = 0$
perpendicular distance $= \frac{11}{\sqrt{5}}$