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Byju's Answer

Standard XII

Mathematics

Equation of a Plane Passing through a Point and Parallel to the Two Given Vectors

Let L1:x-13=y...

Question

Let $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{- 3}$ be a line and $P : 4 x + 3 y + 5 z = 50$ be a plane. $L_{2}$ is the line in the plane $P$ and parallel to $L_{1}$ . If equation of the plane containing both the lines $L_{1}$ and $L_{2}$ and perpendicular to plane $P$ is $a x + b y + 5 z + d = 0,$ then the value of $(a + b + d)$ is

12.0

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Solution

Here, line $L_{2}$ is parallel to $L_{1}$ and lying in the plane $P .$
Now, normal of the plane containing $L_{1}$ and $L_{2}$ both is $⊥$ to the plane $P$ and the lines $L_{1}$ and $L_{2} .$
$∴$ Dr's of normal to the required plane is
$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 3 & 1 & - 3 4 & 3 & 5 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 14^i - 27^j + 5^k$
Thus, equation of the required plane is
$14 (x - 1) - 27 (y - 2) + 5 (z - 3) = 0$
$\Rightarrow 14 x - 27 y + 5 z + 25 = 0$
$\Rightarrow a = 14, b = - 27, d = 25$
Hence, the value of $a + b + d = 14 - 27 + 25 = 12$

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Similar questions

Q. Let

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

be a line and

P : 4 x + 3 y + 5 z - 50 = 0

be a plane.

L_{2}

is the line in the plane

P

and parallel to

L_{1} .

If equation of the plane containing both the lines

L_{1}

and

L_{2}

and perpendicular to plane

P

a x + b y + 5 z + d = 0,

then the value of

a + b + d

Q. Let

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

be a line and

P : 4 x + 3 y + 5 z - 50 = 0

be a plane.

L_{2}

is the line in the plane

P

and parallel to

L_{1} .

If equation of the plane containing both the lines

L_{1}

and

L_{2}

and perpendicular to plane

P

a x + b y + 5 z + d = 0,

then the value of

a + b + d

Q. Let

P

be the plane containing the line

L_{1} : y + z = 2,

x = 0

and is parallel to the line

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

If the distance of the plane

P

from the origin is

d

units, then the value of

3 d^{2}

Q. Assertion :

Consider three planes:

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

Let

L_{1}, L_{2}, L_{3}

be the lines of intersection of the planes

P_{2}

and

P_{3}, P_{3}

and

P_{1}, P_{1}

and

P_{2},

respectively.

At least two of the lines

L_{1}, L_{2}

and

L_{3}

are non-parallel. Reason: The three planes do not have a common point.

Consider the lines $L_{1}$ : $\frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2}$ : $\frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1}$ : $7 x + y + 2 z = 3, P_{2}$ : $3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ .

Match List $1$ with List $2$ and select the correct answer using the code given below the lists:

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Let L1:x−13=y−21=z−3−3 be a line and P:4x+3y+5z=50 be a plane. L2 is the line in the plane P and parallel to L1. If equation of the plane containing both the lines L1 and L2 and perpendicular to plane P is ax+by+5z+d=0, then the value of (a+b+d) is