Question

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. If \(|\vec{a}| = \sqrt{26}\), \(|\vec{b}| = 7\) and \(|\vec{a} \times \vec{b}| = 35\), then \(|\vec{a} \cdot \vec{b}|\) is equal to	I. 4
B. If \(2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(a\hat{i} - 6\hat{j} + 8\hat{k}\) are collinear, then 'a' is equal to	II. 2
C. If \(2\hat{i} + \hat{j} + \hat{k}\) and \(2\hat{i} - 4\hat{j} + \lambda\hat{k}\) are perpendicular, then \(\lambda\) is equal to	III. 7
D. \((\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}\) is equal to	IV. 0

Choose the correct answer from the options given below:

• A. A-III, B-II, C-IV, D-I
• B. A-III, B-I, C-IV, D-II
• C. A-II, B-I, C-III, D-IV
• D. A-I, B-IV, C-II, D-III

Hint:

Use \(|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\), proportional components for collinear and zero dot product for perpendicular.

Answer 1

The Correct Option is B

Step 1: Use the angle for A.
From the given lengths, $\sin\theta = \frac{35}{\sqrt{26} \times 7} = \frac{5}{\sqrt{26}}$. Then $\cos^2\theta = 1 - \frac{25}{26} = \frac{1}{26}$, so $|\cos\theta| = \frac{1}{\sqrt{26}}$.
$|\vec{a} \cdot \vec{b}| = \sqrt{26} \times 7 \times \frac{1}{\sqrt{26}} = 7$. So A is III.

Step 2: Use a scale factor for B.
Set $a\hat{i} - 6\hat{j} + 8\hat{k} = k(2\hat{i} - 3\hat{j} + 4\hat{k})$. The $\hat{j}$ part gives $k = 2$, and the $\hat{k}$ part agrees ($8 = 4k$). Then $a = 2k = 4$. So B is I.

Step 3: Use the dot product for C.
The dot product is $4 - 4 + \lambda$. Setting it to zero gives $\lambda = 0$. So C is IV.

Step 4: Use the triple product for D.
Since $\hat{i}, \hat{j}, \hat{k}$ form a right handed set, $[\hat{i}\ \hat{j}\ \hat{k}] = 1$ and $[\hat{j}\ \hat{k}\ \hat{i}] = 1$. The sum is 2. So D is II.

Step 5: Pick the option.
A-III, B-I, C-IV, D-II is option 2.

Final Answer:
Option 2 is correct. \[ \boxed{\text{Option 2}} \]