Question

$(2\hat{\mathrm{i}} + 6\hat{\mathrm{j}} + 27\hat{\mathrm{k}}) \times (\hat{\mathrm{i}} + \lambda\hat{\mathrm{j}} + \mu\hat{\mathrm{k}}) = \vec{0}$, then $\lambda$ and $\mu$ are respectively

• A. $\frac{17}{2}, 3$
• B. $3, \frac{17}{2}$
• C. $3, \frac{27}{2}$
• D. $\frac{27}{2}, 3$

Hint:

Skip the long determinant grid expansion if the result is zero! Just look at the first component: it drops from $2 \rightarrow 1$ (cut in half). That means every other component must also be cut exactly in half: $\lambda = \frac{6}{2} = 3$ and $\mu = \frac{27}{2}$. It's an instant mental solution!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem states that the vector cross product of two three-dimensional vectors yields the zero vector ($\vec{0}$). We need to solve for the unknown scalar parameters $\lambda$ and $\mu$.

Step 2: Key Formula or Approach:
When the cross product of two non-zero vectors is zero, it implies that the vectors are parallel or collinear. For two parallel vectors, their corresponding $\hat{\mathrm{i}}$, $\hat{\mathrm{j}}$, and $\hat{\mathrm{k}}$ components must be directly proportional: $$ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$

Step 3: Detailed Explanation:
Let's set up the component proportionality ratios from our given vectors: $$ \frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu} $$ This sets up two simple separate equations to solve for each variable:

• Solving for $\lambda$: $$ \frac{2}{1} = \frac{6}{\lambda} \implies 2\lambda = 6 \implies \lambda = 3 $$

• Solving for $\mu$: $$ \frac{2}{1} = \frac{27}{\mu} \implies 2\mu = 27 \implies \mu = \frac{27}{2} $$
Thus, the values are $\lambda = 3$ and $\mu = \frac{27}{2}$.

Step 4: Final Answer:
The values of $\lambda$ and $\mu$ are $3$ and $\frac{27}{2}$ respectively, matching option (C).