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AP DSC School Assistant Math Model Test 2

Section: Subject Specific

Question : 71 of 160

Marks: +1, -0

The vectors

λ\hat{i}+\hat{j}+2\hat{k},\hat{i}+λ\hat{j}-\hat{k}

2\hat{i}-\hat{j}+λ\hat{k}

are coplanar if λ =

Solution:

Concept:

Three vectors are coplanar if their scalar triple product (the determinant of their components) equals zero.

Explanation:

Let the vectors be

\vec{a} = \lambda\hat{i} + \hat{j} + 2\hat{k}

\vec{b} = \hat{i} + \lambda\hat{j} - \hat{k}

\vec{c} = 2\hat{i} - \hat{j} + \lambda\hat{k}

For coplanarity,

[\vec{a},\vec{b},\vec{c}] = 0

, i.e.,

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

Expanding the determinant:

\lambda(\lambda^2 - 1) - 1(\lambda + 2) + 2(-1 - 2\lambda) = 0

\Rightarrow \lambda^3 - \lambda - \lambda - 2 - 2 - 4\lambda = 0

\Rightarrow \lambda^3 - 6\lambda - 4 = 0

Factorising:

(\lambda + 2)(\lambda^2 - 2\lambda - 2) = 0

Thus

\lambda = -2

\lambda = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}

Among the given options, only

\lambda = -2

is listed.

Answer:

\lambda = -2

(Option A)

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