Question

Let  \(λ∈Z ,→a=λ\^i+\^j =/^ k λ ∈\)  and \(\overrightarrow{ b} = 3 \^i− \^j + 2 \^k\) . be a vector such that  ( \(\overrightarrow{ a}+\overrightarrow{ b}+\overrightarrow{ c}\) ) × \(\overrightarrow{ c}=\overrightarrow{ 0},\overrightarrow{ a}.\overrightarrow{ c}\) and  \(\overrightarrow{ b}.\overrightarrow{ C}=-20\)  Then \( | \overrightarrow{c} + ( λ \^i + \^j + \^k ) |  ^2 \) is equal to

• A. 46
• B. 49
• C. 53
• D. 62

Answer 1

The Correct Option is A

Given:

Expanding the equation:

\( (\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0 \)

\( (\vec{a} + \vec{b}) \times \vec{c} = 0 \)

Let:

\( \vec{c} = \alpha(\vec{a} + \vec{b}) = \alpha(\lambda + 3)\hat{i} + \alpha \lambda \hat{k} \)

Using the condition \( \vec{b} \cdot \vec{c} = -20 \):

\( 3\alpha(\lambda + 3) + 2\alpha = -20 \)

Using the condition \( \vec{a} \cdot \vec{c} = -17 \):

\( \alpha(\lambda(\lambda + 3)) - \alpha = -17 \)

From the first equation:

\( \alpha(3\lambda + 9 + 2) = -20 \Rightarrow \alpha(\lambda^2 + 3\lambda - 1) = -17 \)

Simplify and solve:

\( 17(3\lambda + 11) = 20(\lambda^2 + 3\lambda - 1) \)

This simplifies to:

\( 20\lambda^2 + 9\lambda - 207 = 0 \)

Factoring:

\( \lambda = 3 \quad (\lambda \in \mathbb{Z}) \)

Substitute \( \lambda = 3 \) into \( \alpha \):

\( \alpha = -1, \quad \vec{c} = -6\hat{i} + \hat{k} \)

Now, calculate \( \vec{v} = \vec{c} \times (3\hat{i} + \hat{j} + \hat{k}) \):

\( \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & 1 \\ 3 & 1 & 1 \end{vmatrix} \)

Expanding the determinant:

\( \vec{v} = \hat{i}(0 \cdot 1 - (-1) \cdot 1) - \hat{j}(-6 \cdot 1 - 3 \cdot (-1)) + \hat{k}(-6 \cdot 1 - 0 \cdot 3) \)

\( \vec{v} = \hat{i}(1) + \hat{j}(-3) - \hat{k}(-6) = \hat{i} + 3\hat{j} - 6\hat{k} \)

Finally, calculate \( |\vec{v}|^2 \):

\( |\vec{v}|^2 = (1)^2 + (3)^2 + (-6)^2 = 1 + 9 + 36 = 46 \)