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Mathematics Question on Vectors

Let a⃗=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k}, b⃗=3(i^−j^+k^)\vec{b} = 3(\hat{i} - \hat{j} + \hat{k}). Let c⃗\vec{c} be the vector such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b} and a⃗⋅c⃗=3\vec{a} \cdot \vec{c} = 3. Then a⃗⋅((c⃗×b⃗)−b⃗⋅c⃗)\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c}) is equal to:

A

32

B

24

C

20

D

36

Answer

24

Explanation

Solution

Given vectors:
a⃗=i+2j+k,b⃗=3(i−j+k)\vec{a} = i + 2j + k, \quad \vec{b} = 3(i - j + k)

Let c⃗\vec{c} be a vector such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b} and a⃗⋅c⃗=3\vec{a} \cdot \vec{c} = 3. We need to evaluate:

a⃗⋅[(c⃗×b⃗)−b⃗−c⃗]\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right]

Step 1. Expression Simplification: Consider:

a⃗⋅[(c⃗×b⃗)−b⃗−c⃗]=a⃗⋅(c⃗×b⃗)−a⃗⋅b⃗−a⃗⋅c⃗...(i)\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right] = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \quad \text{...(i)}

Step 2. Given Conditions: It is given that:

a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b}

Therefore:

a⃗⋅(c⃗×b⃗)=b⃗⋅b⃗=∣b⃗∣2\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{b} \cdot \vec{b} = |\vec{b}|^2

Calculating the magnitude:

b⃗=3(i−j+k)\vec{b} = 3(i - j + k)

∣b⃗∣2=32[(1)2+(−1)2+(1)2]=27|\vec{b}|^2 = 3^2[(1)^2 + (-1)^2 + (1)^2] = 27

Thus:

a⃗⋅(c⃗×b⃗)=27...(ii)\vec{a} \cdot (\vec{c} \times \vec{b}) = 27 \quad \text{...(ii)}

Step 3. Calculating a⃗⋅b⃗\vec{a} \cdot \vec{b}:

a⃗⋅b⃗=(1)(3)+(2)(−3)+(1)(3)=3−6+3=0...(iii)\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0 \quad \text{...(iii)}

Step 4. Given a⃗⋅c⃗\vec{a} \cdot \vec{c}:

a⃗⋅c⃗=3...(iv)\vec{a} \cdot \vec{c} = 3 \quad \text{...(iv)}

Step 5. Final Calculation: Substituting the values from (ii), (iii), and (iv) into (i):

a⃗⋅[(c⃗×b⃗)−b⃗−c⃗]=27−0−3=24\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right] = 27 - 0 - 3 = 24