Test Index

Vector Algebra Part 3

Section: Mathematics

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Question : 11 of 72

Marks: +1, -0

Let

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

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

(\vec{a} - \vec{b}) \cdot \vec{d}

[23 Jan 2026 Shift 1]

2
-2
4
-4

Solution:

Concept:

Use scalar triple product properties and vector triple product identity to simplify the dot product.

Explanation:

Given

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

and

\vec{b} = \hat{i} - \hat{j} - 3\hat{k}

.

First compute:

\vec{a} \cdot \vec{b} = (-1)(1) + (1)(-1) + (2)(-3) = -8

.

|\vec{a}|^2 = \vec{a} \cdot \vec{a} = 1+1+4 = 6

.

|\vec{b}|^2 = \vec{b} \cdot \vec{b} = 1+1+9 = 11

.

Now

(\vec{a} - \vec{b}) \cdot \vec{d} = \vec{a} \cdot \vec{d} - \vec{b} \cdot \vec{d}

.

Since

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

and

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

, we have

\vec{a} \cdot \vec{d} = [\vec{a} \, \vec{c} \, \vec{a}] = 0

.

Thus

(\vec{a} - \vec{b}) \cdot \vec{d} = - \vec{b} \cdot \vec{d}

.

Now

\vec{b} \cdot \vec{d} = \vec{b} \cdot ((\vec{a} \times \vec{b}) \times \vec{a})

.

Using the triple product identity

(\vec{p} \times \vec{q}) \times \vec{r} = (\vec{p} \cdot \vec{r})\vec{q} - (\vec{q} \cdot \vec{r})\vec{p}

,

(\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{a})\vec{a} = |\vec{a}|^2 \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}

.

Hence

\vec{b} \cdot \vec{d} = \vec{b} \cdot (|\vec{a}|^2 \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}) = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2

.

Substitute values:

= 6 \times 11 - (-8)^2 = 66 - 64 = 2

.

Therefore

(\vec{a} - \vec{b}) \cdot \vec{d} = - (2) = -2

.

Answer:

-2

(Option B)

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