Question 96: Motion in 2D & Vectors

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

Accepted Answer

We are given vectors:
$ \vec{P}=b \hat{i}+6 \hat{j}+\hat{k}, \quad \vec{Q}=\hat{i}-a \hat{j}+4 \hat{k}. $
Step 1: Condition for perpendicularity
$ \vec{P}\cdot \vec{Q}=0 $
$ (b)(1) + (6)(-a) + (1)(4)= 0 $
$ b - 6a + 4 = 0 \quad \Rightarrow \quad b - 6a = -4 \quad \Rightarrow \quad b = 6a - 4 $
Step 2: Given relation
$ 3b - a = 5 $
Substitute $b = 6a - 4$:
$ 3(6a - 4) - a = 5 $
$ 18a - 12 - a = 5 $
$ 17a - 12 = 5 $
$ 17a = 17 \quad \Rightarrow \quad a = 1 $
Step 3: Solve for $b$
$ b = 6(1) - 4 = 2 $
Ã¢ÂÂ Final Answer:
$ a = 1, \; b = 2 $
Correct Option: B

Answer

$\mathrm{a}=2, \mathrm{~b}=10$

Answer

$\mathrm{a}=1, \mathrm{~b}=2$

Answer

$\mathrm{a}=2, \mathrm{~b}=3$

Answer

$\mathrm{a}=4, \mathrm{~b}=3$

Motion in 2D & Vectors

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

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