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JEE Main 2023 · 25 Jan · Shift 2

JEE Main 2023 Mathematics question on Vector Algebra

If the four points, whose position vectors are 3i^−4j^+2k^3\hat{i} - 4\hat{j} + 2\hat{k}, i^+2j^−k^\hat{i} + 2\hat{j} - \hat{k}, −2i^−j^+3k^-2\hat{i} - \hat{j} + 3\hat{k} and 5i^−2aj^+4k^5\hat{i} - 2a\hat{j} + 4\hat{k} are coplanar, then aa is equal to
  1. A.
    7317\dfrac{73}{17}
  2. B.
    −7317-\dfrac{73}{17}
  3. C.
    10717\dfrac{107}{17}
  4. D.
    −10717-\dfrac{107}{17}
Show answer and solution

Correct option: A

Solution

Let A = (3,-4,2), B = (1,2,-1), C = (-2,-1,3), D = (5,-2a,4).

AB = (-2,6,-3), AC = (-5,3,1), AD = (2,-2a+4,2).

For coplanarity, the scalar triple product [AB, AC, AD] = 0.

|−2 6 −3|
|−5 3 1| = 0
| 2 −2a+4 2|

Expanding: -2(3·2 - 1·(-2a+4)) - 6(-5·2 - 1·2) + (-3)(-5(-2a+4) - 3·2)

= -2(6 - (-2a+4)) - 6(-10-2) + (-3)(10a - 20 - 6)

= -2(6 + 2a - 4) - 6(-12) + (-3)(10a - 26)

= -2(2 + 2a) + 72 - 30a + 78

= -4 - 4a + 72 - 30a + 78

= 146 - 34a = 0

So a = 146/34 = 73/17.

[JEE Main 2023 · 25 Jan · Shift 2]

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