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Mathematics Question on Vector Algebra

Let a⃗\vec{a} = 2i^\hat{i} + 5j^\hat{j} - k^\hat{k}, b⃗\vec{b} = 2i^\hat{i} - 2j^\hat{j} + 2k^\hat{k}
and c⃗\vec{c} be three vectors such that
(c⃗\vec{c} + i^\hat{i}) ×\times (a⃗\vec{a} + b⃗\vec{b} + i^\hat{i}) = a⃗\vec{a} ×\times (c⃗\vec{c} + i^)\hat{i}) . a⃗\vec{a}.c⃗\vec{c} = -29,)
then c⃗\vec{c}.(-2i^\hat{i} + j^\hat{j} + k^\hat{k}) is equal to :

A

10

B

5

C

15

D

12

Answer

5

Explanation

Solution

Let us analyze the given conditions step by step.

Step 1: Represent the vector v⃗\vec{v}

Define:

v⃗=a⃗+b⃗+i^\vec{v} = \vec{a} + \vec{b} + \hat{i}.

Substitute the given values of a⃗\vec{a} and b⃗\vec{b}:

v⃗=(2i^+5j^−k^)+(2i^−2j^+2k^)+i^\vec{v} = (2\hat{i} + 5\hat{j} - \hat{k}) + (2\hat{i} - 2\hat{j} + 2\hat{k}) + \hat{i}.

Simplify:

v⃗=(2+2+1)i^+(5−2)j^+(−1+2)k^=5i^+3j^+k^\vec{v} = (2 + 2 + 1)\hat{i} + (5 - 2)\hat{j} + (-1 + 2)\hat{k} = 5\hat{i} + 3\hat{j} + \hat{k}.

Step 2: Represent c⃗+i^\vec{c} + \hat{i} as p⃗\vec{p}

Define:

p⃗=c⃗+i^\vec{p} = \vec{c} + \hat{i}.

Step 3: Apply the cross-product condition

The condition is:

p⃗×v⃗=a⃗×p⃗\vec{p} \times \vec{v} = \vec{a} \times \vec{p}.

Rearrange:

p⃗×v⃗−p⃗×a⃗=0⃗\vec{p} \times \vec{v} - \vec{p} \times \vec{a} = \vec{0}.

Using the distributive property of the cross product:

p⃗×(v⃗−a⃗)=0⃗\vec{p} \times (\vec{v} - \vec{a}) = \vec{0}.

Thus, p⃗\vec{p} must be parallel to v⃗−a⃗\vec{v} - \vec{a}, which implies:

p⃗=λ(v⃗−a⃗)\vec{p} = \lambda (\vec{v} - \vec{a}), where λ\lambda is a scalar.

Step 4: Substitute v⃗−a⃗\vec{v} - \vec{a}

Calculate v⃗−a⃗\vec{v} - \vec{a}:

v⃗−a⃗=(5i^+3j^+k^)−(2i^+5j^−k^)\vec{v} - \vec{a} = (5\hat{i} + 3\hat{j} + \hat{k}) - (2\hat{i} + 5\hat{j} - \hat{k}).

Simplify:

v⃗−a⃗=(5−2)i^+(3−5)j^+(1−(−1))k^=3i^−2j^+2k^\vec{v} - \vec{a} = (5 - 2)\hat{i} + (3 - 5)\hat{j} + (1 - (-1))\hat{k} = 3\hat{i} - 2\hat{j} + 2\hat{k}.

Thus:

p⃗=λ(3i^−2j^+2k^)\vec{p} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k}).

Step 5: Use the dot-product condition

The condition a⃗⋅c⃗=−29\vec{a} \cdot \vec{c} = -29 can be written as:

a⃗⋅(p⃗−i^)=−29\vec{a} \cdot (\vec{p} - \hat{i}) = -29.

Substitute p⃗=λ(3i^−2j^+2k^)\vec{p} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k}):

a⃗⋅(λ(3i^−2j^+2k^)−i^)=−29\vec{a} \cdot (\lambda (3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}) = -29.

Expand:

a⃗⋅(λ3i^−λ2j^+λ2k^−i^)=−29\vec{a} \cdot (\lambda 3\hat{i} - \lambda 2\hat{j} + \lambda 2\hat{k} - \hat{i}) = -29.

Substitute a⃗=2i^+5j^−k^\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}:

a⃗⋅(3λi^−2λj^+2λk^−i^)=−29\vec{a} \cdot (3\lambda \hat{i} - 2\lambda \hat{j} + 2\lambda \hat{k} - \hat{i}) = -29.

Simplify:

(2)(3λ)+(5)(−2λ)+(−1)(2λ)−(2)(1)=−29(2)(3\lambda) + (5)(-2\lambda) + (-1)(2\lambda) - (2)(1) = -29.

Thus:

6λ−10λ−2λ−2=−296\lambda - 10\lambda - 2\lambda - 2 = -29.

Solve for λ\lambda:

−6λ−2=−29  ⟹  −6λ=−27  ⟹  λ=276=−12-6\lambda - 2 = -29 \implies -6\lambda = -27 \implies \lambda = \frac{27}{6} = -\frac{1}{2}.

Step 6: Compute c⃗\vec{c}

Substitute c⃗=p⃗−i^\vec{c} = \vec{p} - \hat{i}:

c⃗=λ(3i^−2j^+2k^)−i^\vec{c} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}.

Simplify:

c⃗=−12(3i^−2j^+2k^)−i^\vec{c} = -\frac{1}{2}(3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}.

c⃗=−32i^+j^−k^−i^\vec{c} = -\frac{3}{2}\hat{i} + \hat{j} - \hat{k} - \hat{i}.

Combine terms:

c⃗=−52i^+j^−k^\vec{c} = -\frac{5}{2}\hat{i} + \hat{j} - \hat{k}.