Let {a}=2 {i}-3 {j}+{k}, {b}=3 {i}+2 {j}+2 {k}, {c}=4 {i}-3 {j}+{k}, then the vectors {a}, {b}, {c} are

Correct Answer is : Linearly independent

Correct Answer is : Linearly independent

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 140 of 150

Marks: +1, -0

Let

\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \quad \vec{b}=3 \hat{i}+2 \hat{j}+2 \hat{k}, \quad \vec{c}=4 \hat{i}-3 \hat{j}+\hat{k}

\vec{a}, \quad \vec{b}, \quad \vec{c}

are

Solution:

Concept:

For three vectors in space, the scalar triple product

[\vec{a}\ \vec{b}\ \vec{c}]

decides whether they are linearly independent or coplanar.

Explanation:

[\vec{a}\ \vec{b}\ \vec{c}] \neq 0

, then the vectors are linearly independent and not coplanar.

Given:

\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}

\vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k}

\vec{c} = 4\hat{i} - 3\hat{j} + \hat{k}

So,

[\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 2 & 2 \\ 4 & -3 & 1 \end{vmatrix}

Expanding along the first row:

= 2\left(2 \cdot 1 - 2 \cdot (-3)\right) - (-3)\left(3 \cdot 1 - 2 \cdot 4\right) + 1\left(3 \cdot (-3) - 2 \cdot 4\right)

= 2(2 + 6) + 3(3 - 8) + 1(-9 - 8)

= 2(8) + 3(-5) - 17

= 16 - 15 - 17

= -16

Since

[\vec{a}\ \vec{b}\ \vec{c}] = -16 \neq 0

, the vectors

\vec{a}, \vec{b}, \vec{c}

are linearly independent.

Answer:

Option C: Linearly independent

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