Q. 263.7( 3 Votes )

# Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0. Also find the distance of the plane obtained above, from the origin.

**OR**

Find the distance of the point (2, 12, 5) from the point of intersection of the line and the plane **[CBSE 2014]**

**[CBSE 2014]**

Answer :

Let P_{1} be the plane x + y + z = 1 and P_{2} be the plane 2x + 3y + 4z = 5 and P_{3} be the plane x - y + z = 0

The plane passing through line of intersection of P_{1} and P_{2} is given by P_{1} + λP_{2} = 0

⇒ x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0

⇒ (2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1 = 0

Let this be the plane P which is required

⇒ P = (2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1

The direction of normal to a plane with equation ax + by + cz + d = 0 is <a, b, c> or

Hence the direction of normal to plane P is

It is given that the required plane P is perpendicular to plane P_{3}

As the planes P and P_{3} are perpendiculars, their normal vectors are also perpendicular

The direction of normal to plane P_{3} is

Take dot product of normal of plane P and normal of plane P_{3}

⇒ 2λ + 1 – 3λ – 1 + 4λ + 1 = 0

⇒ 3λ + 1 = 0

Put value of lambda in P

⇒ (2λ + 1)x + (3λ + 1)y + (4λ + 1)z – 5λ – 1 = 0

Multiply by 3

⇒ x – z + 2 = 0

Distance of plane ax + by + cz + d = 0 from a given point (x_{1}, y_{1}, z_{1}) is given by

Hence distance of plane x – z + 2 = 0 from origin is

⇒ distance = √2 units

Hence equation of required plane is x – z + 2 = 0 and its distance from origin is √2 units

**OR**

To find the distance of the point (2, 12, 5) from the point of intersection of line and plane, we must first find out the point of intersection of line and plane

The equation of a line given

Comparing with the standard form where is point on line and is direction

and

The equation of a line in cartesian form is where (a_{1}, a_{2}, a_{3}) is the point on line and <b_{1}, b_{2}, b_{3}> is the direction

Hence equation of line in cartesian form

⇒ x – 2 = 3k, y + 4 = 4k, z – 2 = 2k

⇒ x = 3k + 2, y = 4k – 4, z = 2k + 2

Hence general point on line is given by (3k + 2, 4k – 4, 2k + 2)

The equation of plane is

Put to get the cartesian form

⇒ x – 2y + z = 0

As the line intersects the plane hence, we will put the general point on the line in the equation of plane so that we will get a specific value of k for which the point will lie on both plane and line which means the intersecting point of plane and line

⇒ (3k + 2) – 2(4k – 4) + 2k + 2 = 0

⇒ 5k + 4 – 8k + 8 = 0

⇒ -3k + 12 = 0

⇒ 3k = 12

⇒ k = 4

Hence for k = 4 the general point on line is intersecting point of line and plane

Hence the point is (3(4) + 2, 4(4) – 4, 2(4) + 2) = (14, 12, 10)

Now we have to find the distance between (2, 12, 5) and (14, 12, 10)

Using distance formula

**⇒** **distance = 13 units**

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**OR**

Find the distance of the point (2, 12, 5) from the point of intersection of the line and the plane

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