Question

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

• A. 10
• B. 5
• C. 15
• D. 12

Answer 1

The Correct Option is B

Let us analyze the given conditions step by step.

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

Define:

$\vec{v} = \vec{a} + \vec{b} + \hat{i}$.

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

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

Simplify:

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

Step 2: Represent $\vec{c} + \hat{i}$ as $\vec{p}$

Define:

$\vec{p} = \vec{c} + \hat{i}$.

Step 3: Apply the cross-product condition

The condition is:

$\vec{p} \times \vec{v} = \vec{a} \times \vec{p}$.

Rearrange:

$\vec{p} \times \vec{v} - \vec{p} \times \vec{a} = \vec{0}$.

Using the distributive property of the cross product:

$\vec{p} \times (\vec{v} - \vec{a}) = \vec{0}$.

Thus, $\vec{p}$ must be parallel to $\vec{v} - \vec{a}$, which implies:

$\vec{p} = \lambda (\vec{v} - \vec{a})$, where $\lambda$ is a scalar.

Step 4: Substitute $\vec{v} - \vec{a}$

Calculate $\vec{v} - \vec{a}$:

$\vec{v} - \vec{a} = (5\hat{i} + 3\hat{j} + \hat{k}) - (2\hat{i} + 5\hat{j} - \hat{k})$.

Simplify:

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

Thus:

$\vec{p} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k})$.

Step 5: Use the dot-product condition

The condition $\vec{a} \cdot \vec{c} = -29$ can be written as:

$\vec{a} \cdot (\vec{p} - \hat{i}) = -29$.

Substitute $\vec{p} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k})$:

$\vec{a} \cdot (\lambda (3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}) = -29$.

Expand:

$\vec{a} \cdot (\lambda 3\hat{i} - \lambda 2\hat{j} + \lambda 2\hat{k} - \hat{i}) = -29$.

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

$\vec{a} \cdot (3\lambda \hat{i} - 2\lambda \hat{j} + 2\lambda \hat{k} - \hat{i}) = -29$.

Simplify:

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

Thus:

$6\lambda - 10\lambda - 2\lambda - 2 = -29$.

Solve for $\lambda$:

$-6\lambda - 2 = -29 \implies -6\lambda = -27 \implies \lambda = \frac{27}{6} = -\frac{1}{2}$.

Step 6: Compute $\vec{c}$

Substitute $\vec{c} = \vec{p} - \hat{i}$:

$\vec{c} = \lambda (3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}$.

Simplify:

$\vec{c} = -\frac{1}{2}(3\hat{i} - 2\hat{j} + 2\hat{k}) - \hat{i}$.

$\vec{c} = -\frac{3}{2}\hat{i} + \hat{j} - \hat{k} - \hat{i}$.

Combine terms:

$\vec{c} = -\frac{5}{2}\hat{i} + \hat{j} - \hat{k}$.