Multiple choice

If $\vec {a}=2\hat {i}-\hat {j}+\hat {k},:\vec {b}=\hat {i}+2\hat {j}-3\hat {k},:\ \vec {c}=3\hat {i}+\mu\hat {j}+5\hat {k}$ are coplanar, then $\mu$ is a root of the equation

  1. $x^{2}+3x=4$
  2. $x^{2}+2x=6$
  3. $x^{2}+3x=6$

  4. None of these
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A Correct answer
Explanation

Three vectors are coplanar if their scalar triple product is zero. The determinant of the matrix formed by the components is 0. Solving the determinant yields a quadratic equation in mu, which matches the form x^2 + 3x = 4.

AI explanation

For the vectors to be coplanar, their scalar triple product must be zero, so we evaluate the determinant of their components: (2, -1, 1), (1, 2, -3), and (3, mu, 5). Expanding this determinant gives 2(10 + 3mu) + 1(5 + 9) + 1(mu - 6) = 0, which simplifies to 7mu + 28 = 0. This means mu is -4, making it a root of the equation x squared + 3x = 4.

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