Question

The value of $\lambda$ for which the vectors $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal is

• A. $\frac{5}{2}$
• B. $\frac{-5}{2}$
• C. $\frac{2}{5}$
• D. $\frac{-2}{5}$

Hint:

"Orthogonal" is just a formal term for "perpendicular." Whenever you see it in a vector problem, immediately write down $\vec{u} \cdot \vec{v} = 0$.

Answer 1

The Correct Option is B

To determine the value of \(\lambda\) for which the vectors \(\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\) are orthogonal, we need to use the condition for orthogonality of two vectors.

Two vectors are orthogonal if their dot product is zero. The dot product of vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\(\vec{a} \cdot \vec{b} = (2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})\)

We calculate each component of the dot product:

• \(2\hat{i} \cdot \hat{i} = 2\)
• \(\lambda\hat{j} \cdot 2\hat{j} = 2\lambda\)
• \(\hat{k} \cdot 3\hat{k} = 3\)

Putting these together, the dot product is:

\(2 + 2\lambda + 3 = 0\)

Simplifying the equation:

\(2\lambda + 5 = 0\)

Solving for \(\lambda\):

\(2\lambda = -5\) 
\(\lambda = \frac{-5}{2}\)

Hence, the value of \(\lambda\) for which the vectors are orthogonal is \(\frac{-5}{2}\).

Therefore, the correct answer is:

$\frac{-5}{2}$