Question

If \(\vec{P}=b \hat{i}+6 \hat{j}+\hat{k} \quad\) and \(\quad \vec{Q}=\hat{i}-a \hat{j}+4 \hat{k} \quad\) are perpendicular to each other, also \(3 \mathrm{~b}-\mathrm{a}=5\). The value of \(a\) and \(b\) is

• A. \(\mathrm{a}=2,\text{ }\mathrm{b}=10\)
• B. \(\mathrm{a}=1,\text{ }\mathrm{b}=2\)
• C. \(\mathrm{a}=2,\text{ }\mathrm{b}=3\)
• D. \(\mathrm{a}=4,\text{ }\mathrm{b}=3\)

Hint:

Perpendicular vectors have zero dot product. Combine that equation with the given linear relation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Two vectors are perpendicular when their dot product is zero.

Step 2: Dot Product
\(\vec{P}\cdot\vec{Q} = b(1) + 6(-a) + 1(4) = b - 6a + 4 = 0\)
So \(b = 6a - 4\).

Step 3: Use the Given Relation
Put \(b = 6a - 4\) in \(3b - a = 5\):
\(18a - 12 - a = 5\)
\(17a = 17\), so \(a = 1\).

Step 4: Find b
\(b = 6(1) - 4 = 2\)

Final Answer:
The values are a equal to 1 and b equal to 2. \[ \boxed{\text{(B) }a = 1,\ b = 2} \]