Let a = 2i - 3j + k, b = 3i + 2j + 5k and a vector c be such that (a - c) x b = -18i - 3j + 12k and a . c = 3. If b x c = d, then |a . d| is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2025

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Home Maths Vector AlgebraQuestion #518

Let

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

\vec{b} = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}

and a vector

\vec{c}

be such that

(\vec{a} - \vec{c}) \times \vec{b} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}

and

\vec{a} \cdot \vec{c} = 3

. If

\vec{b} \times \vec{c} = \vec{d}

, then

| \vec{a} \cdot \vec{d} |

is equal to :

Clue 1 of 3: Vector Properties

Recognize that the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ can be simplified using properties of vector operations.

🧠 Full Solution Path

Step 1: Expand the given vector equation✦ Active

The given equation is $(\vec{a} - \vec{c}) \times \vec{b} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ . Using the distributive property of the cross product, this becomes:

\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}

Since $\vec{c} \times \vec{b} = - (\vec{b} \times \vec{c})$ , and we are given $\vec{d} = \vec{b} \times \vec{c}$ , we can rewrite the equation as:

\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}

Substituting $\vec{d} = \vec{b} \times \vec{c}$ :

\vec{a} \times \vec{b} + \vec{d} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}

Step 2: Utilize the scalar triple product property○ Expand

We need to find $| \vec{a} \cdot \vec{d} |$ . To do this, take the dot product of the equation from Step 1 with vector $\vec{a}$ :

\vec{a} \cdot (\vec{a} \times \vec{b} + \vec{d}) = \vec{a} \cdot (- 18 \hat{i} - 3 \hat{j} + 12 \hat{k})

Using the distributive property of the dot product:

\vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot \vec{d} = \vec{a} \cdot (- 18 \hat{i} - 3 \hat{j} + 12 \hat{k})

We know that the scalar triple product $\vec{a} \cdot (\vec{a} \times \vec{b})$ is zero because two vectors are identical. Thus, the equation simplifies to:

\vec{a} \cdot \vec{d} = \vec{a} \cdot (- 18 \hat{i} - 3 \hat{j} + 12 \hat{k})

💡 Teacher's Secret Hint

The information $\vec{a} \cdot \vec{c} = 3$ is extraneous for this particular solution path.

Step 3: Calculate the dot product and magnitude○ Expand

Given $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}$ . Now, calculate the dot product on the right-hand side:

\vec{a} \cdot \vec{d} = (2) (- 18) + (- 3) (- 3) + (1) (12)

\vec{a} \cdot \vec{d} = - 36 + 9 + 12

\vec{a} \cdot \vec{d} = - 36 + 21

\vec{a} \cdot \vec{d} = - 15

Finally, we need to find the magnitude of this scalar product:

| \vec{a} \cdot \vec{d} | = | - 15 | = 15

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Maths Question 16 – JEE-MAIN 2025

Clue 1 of 3: Vector Properties

🧠 Full Solution Path