Question

Find the vector equation of the plane which passes through the points 2i + 4j + 2k,21 + 3j + 5k and parallel to the vector 3i - 2j+k. Also find the point where this plane meets theline joining the points 2i + % +3% and 4i - 2j + 3k.Can anyone answer dis?​

Answer 1

Answer:

i dont know

Explanation:

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Answer 2

The point is P ( -14/17, 89/17, 3)

Let the resultant point be P.

It is given that the plane passes through the points 2i + 4j + 2k (a) , 2i + 3j + 5k (b) and is parallel to the vector 3i - 2j + k (c).

Now let us find the vector equation of the plane.

r = (1 − s)a + sb + tc

where,

a = 2 i + 4 j + 2k, b = 2 i + 3 j + 5k and c = 3i − 2 j + k.

So, the equation of the plane will be :

r = (1 − s)(2 i + 4 j + 2k) + s(2 i + 3 j + 5k) + t(3t − 2 j + k)

= [2 − 2s + 2s + 3t] i + [4 − 4s + 3s − 2t] j + [2 − 2s + 5s + t] k

= [3t + 2] i + [4 − s − 2t] j + [2 + 3 j + t] k ...(1)

Let us consider the equation of the line to be

r = (1 − p) a + p b

where p is a scalar.

r = (1 − p)(2 i + j + 3k) + p(4 i − 2 j + 3k)

= (2 − 2p + 4p) i +(1 − p − 2p) j +(3 − 3p +3p) k

= [2 + 2p] i + [1 − 3p] j + 3 k     ...(2)

At the point of intersection, we equate (1) = (2). So,

(3t + 2) i + (4 − s − 2t) j + (2 + 3s + t) k = (2 + 2p) i + (1 − 3p) j + 3k

⇒ 3t + 2 = 2p + 2

⇒ 3t- 2p = 0     ...(3)

⇒ 4 − s − 2t = 1 − 3p

⇒ 2t + s − 3p = 3 ...(4)

⇒ 2 + 3s + t = 3

⇒ 3s + t = 1 ...(5)

Solving (4) and (5), we get,

5t – 9p = 8 ...(6)

Subtracting (6) from (3) we get,

15t – 10 p = 0                                [3t- 2p = 0]

⇒ 15t – 27p = 24

⇒ 17p = –24

⇒ p = -24/17

Putting the value of p in eqn.(2),

r = (2 + 2×(-24/17)) i + (1 - 3×(-24/17)) j + 3k

= (-14/17)i + (89/17)j + 3 k

So, the resultant point is :

P ( -14/17,  89/17,  3)