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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Vector Algebra

Let a=i^−j^\mathbf{a} = \hat{i} - \hat{j}, b=i^+j^+k^\mathbf{b} = \hat{i} + \hat{j} + \hat{k} and c\mathbf{c} be a vector such that a×c+b=0\mathbf{a} \times \mathbf{c} + \mathbf{b} = \mathbf{0} and a⋅c=4\mathbf{a} \cdot \mathbf{c} = 4, then ∣c∣2|\mathbf{c}|^2 is equal to
  1. A.
    8
  2. B.
    192\dfrac{19}{2}
  3. C.
    9
  4. D.
    172\dfrac{17}{2}
Show answer and solution

Correct option: B

Solution

Let c = (x, y, z). From a·c = 4: x - y = 4.

From a×c + b = 0, we get a×c = -b = (-1, -1, -1).

a×c = (i-j)×(xi+yj+zk) = (i×yj) + (i×zk) + (-j×xi) + (-j×zk)
= yk̂ - zĵ - x(-k̂) - (-xi×... let me compute directly:

a×c = |i j k|
|1 -1 0|
|x y z|

= i((-1)z - 0·y) - j(1·z - 0·x) + k(1·y - (-1)x)
= i(-z) - j(z) + k(y+x)
= (-z, -z, x+y)

Setting equal to (-1, -1, -1):
-z = -1 → z = 1
-z = -1 → consistent
x + y = -1

Combined with x - y = 4:
Adding: 2x = 3 → x = 3/2, y = -5/2.

|c|² = x² + y² + z² = (9/4) + (25/4) + 1 = 34/4 + 1 = 17/2 + 2/2... wait: 34/4 + 4/4 = 38/4 = 19/2.

So |c|² = (3/2)² + (-5/2

[JEE Main 2019 · 9 Jan · Shift I]

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