Let a = 4i - j + 3k, b = 10i + 2j - k and a vector c be such that 2(a x b) + 3(b x c) = 0. If a . c = 15, then c . (i + j - 3k) is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2026

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Home Maths Vector AlgebraQuestion #513

Let

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

\vec{b} = 10 \hat{i} + 2 \hat{j} - \hat{k}

and a vector

\vec{c}

be such that

2 (\vec{a} \times \vec{b}) + 3 (\vec{b} \times \vec{c}) = \vec{0}

. If

\vec{a} \cdot \vec{c} = 15

, then

\vec{c} \cdot (\hat{i} + \hat{j} - 3 \hat{k})

is equal to :

Clue 1 of 3: Vector Cross Product Properties

The given vector equation $2 (\vec{a} \times \vec{b}) + 3 (\vec{b} \times \vec{c}) = \vec{0}$ can be rearranged using the property $\vec{X} \times \vec{Y} = - (\vec{Y} \times \vec{X})$ to reveal a relationship of parallelism between vectors.

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Step 1: Simplify the vector equation and express

\vec{c}

✦ Active

The given vector equation is $2 (\vec{a} \times \vec{b}) + 3 (\vec{b} \times \vec{c}) = \vec{0}$ . Using the property $\vec{b} \times \vec{c} = - (\vec{c} \times \vec{b})$ , we can rewrite it as $2 (\vec{a} \times \vec{b}) - 3 (\vec{c} \times \vec{b}) = \vec{0}$ . This simplifies to $(2 \vec{a} - 3 \vec{c}) \times \vec{b} = \vec{0}$ . This implies that the vector $(2 \vec{a} - 3 \vec{c})$ is parallel to $\vec{b}$ . Therefore, we can write $2 \vec{a} - 3 \vec{c} = λ \vec{b}$ for some scalar $λ$ . From this, we express $\vec{c}$ as:

\vec{c} = \frac{2}{3} \vec{a} - \frac{λ}{3} \vec{b}

Step 2: Determine the scalar

λ

and find

\vec{c}

○ Expand

We are given $\vec{a} \cdot \vec{c} = 15$ . Substitute the expression for $\vec{c}$ from Step 1:

\vec{a} \cdot (\frac{2}{3} \vec{a} - \frac{λ}{3} \vec{b}) = 15 ⟹ \frac{2}{3} (\vec{a} \cdot \vec{a}) - \frac{λ}{3} (\vec{a} \cdot \vec{b}) = 15

Calculate the required dot products:

| \vec{a} |^{2} = (4)^{2} + (- 1)^{2} + (3)^{2} = 16 + 1 + 9 = 26

\vec{a} \cdot \vec{b} = (4) (10) + (- 1) (2) + (3) (- 1) = 40 - 2 - 3 = 35

Substitute these values into the equation for $λ$ :

\frac{2}{3} (26) - \frac{λ}{3} (35) = 15 ⟹ 52 - 35 λ = 45 ⟹ 35 λ = 7 ⟹ λ = \frac{1}{5}

Now, substitute $λ = \frac{1}{5}$ back into the expression for $\vec{c}$ :

\vec{c} = \frac{2}{3} \vec{a} - \frac{1}{15} \vec{b} = \frac{2}{3} (4 \hat{i} - \hat{j} + 3 \hat{k}) - \frac{1}{15} (10 \hat{i} + 2 \hat{j} - \hat{k})

\vec{c} = (\frac{8}{3} - \frac{10}{15}) \hat{i} + (- \frac{2}{3} - \frac{2}{15}) \hat{j} + (2 + \frac{1}{15}) \hat{k}

\vec{c} = (\frac{40 - 10}{15}) \hat{i} + (\frac{- 10 - 2}{15}) \hat{j} + (\frac{30 + 1}{15}) \hat{k} = 2 \hat{i} - \frac{4}{5} \hat{j} + \frac{31}{15} \hat{k}

💡 Teacher's Secret Hint

Ensure careful calculation of dot products and fractions to avoid arithmetic errors.

Step 3: Calculate the final required dot product○ Expand

Finally, we need to calculate $\vec{c} \cdot (\hat{i} + \hat{j} - 3 \hat{k})$ :

\vec{c} \cdot (\hat{i} + \hat{j} - 3 \hat{k}) = (2 \hat{i} - \frac{4}{5} \hat{j} + \frac{31}{15} \hat{k}) \cdot (\hat{i} + \hat{j} - 3 \hat{k})

= (2) (1) + (- \frac{4}{5}) (1) + (\frac{31}{15}) (- 3)

= 2 - \frac{4}{5} - \frac{93}{15} = 2 - \frac{4}{5} - \frac{31}{5}

= 2 - \frac{35}{5} = 2 - 7 = - 5

The value is $- 5$ , which corresponds to option 2.

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Maths Question 14 – JEE-MAIN 2026

Clue 1 of 3: Vector Cross Product Properties

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