Question

If the vectors \(2\hat{i}-3\hat{j}+4\hat{k},2\hat{i}+\hat{j}-\hat{k}\) and \(\lambda\hat{i}-\hat{j}+2\hat{k}\) are coplanar, then the value of λ is

• A. 6
• B. -5
• C. -6
• D. 5

Answer 1

The Correct Option is A

For three vectors to be coplanar, their scalar triple product must be zero. That is, the determinant of the matrix formed by the components of the vectors must be zero.

\(\begin{vmatrix} 2 & -3 & 4 \\ 2 & 1 & -1 \\ \lambda & -1 & 2 \end{vmatrix} = 0\)

Expanding the determinant:

\(2(1 \cdot 2 - (-1)(-1)) - (-3)(2 \cdot 2 - (-1)(\lambda)) + 4(2(-1) - 1(\lambda)) = 0\)

\(2(2 - 1) + 3(4 + \lambda) + 4(-2 - \lambda) = 0\)

\(2(1) + 12 + 3\lambda - 8 - 4\lambda = 0\)

\(2 + 12 - 8 + 3\lambda - 4\lambda = 0\)

\(6 - \lambda = 0\)

\(\lambda = 6\)

Therefore, the value of \(\lambda\) is 6.

Thus, the correct option is (A) 6.

Answer 2

For vectors to be coplanar, their scalar triple product must be zero. The scalar triple product is given by:

$$ \begin{vmatrix} 2 & -3 & 4 \\ 2 & 1 & -1 \\ \lambda & -1 & 2 \end{vmatrix} = 0. $$

Expanding the determinant:

$$ 2(2-(-1)) - (-3)(4+\lambda) + 4(-2-\lambda) = 0. $$

Simplify each term:

$$ 2(1) + 3(4+\lambda) + 4(-2-\lambda) = 0. $$ $$ 2 + 12 + 3\lambda - 8 - 4\lambda = 0. $$ $$ 6 - \lambda = 0. $$ $$ \lambda = 6. $$