If two vectors \overrightarrow P = \widehat i + 2m\widehat j + m\widehat k and \overrightarrow Q = 4\widehat i - 2\widehat j + m\widehat k are perpendicular to each other. Then, the value of m will be :

Vector Algebra question

2023 · 24 Jan · Shift 2 · Q62

2023 · 24 Jan · Shift 2 · Q62

View written solutionFree

Correct answer: D

For two vectors to be perpendicular, their dot product must be zero.

Given: $\vec P = \hat i + 2m\hat j + m\hat k$ $\vec Q = 4\hat i - 2\hat j + m\hat k$
Write their components: $\vec P = (1,\, 2m,\, m)$ $\vec Q = (4,\, -2,\, m)$
Compute the dot product: $\vec P \cdot \vec Q = (1)(4) + (2m)(-2) + (m)(m)$ $= 4 - 4m + m^2$
Since the vectors are perpendicular: $4 - 4m + m^2 = 0$ $m^2 - 4m + 4 = 0$ $(m-2)^2 = 0$
Therefore, $m = 2$
Check options:
- A: $-1$ ❌
- B: $3$ ❌
- C: $1$ ❌
- D: $2$ ✅

Hence, the correct answer is Option D.

2023 · 24 Jan · Shift 2 · Q62

2023 · 24 Jan · Shift 2 · Q62

View written solutionFree

Correct answer: D

For two vectors to be perpendicular, their dot product must be zero.

Given: $\vec P = \hat i + 2m\hat j + m\hat k$ $\vec Q = 4\hat i - 2\hat j + m\hat k$
Write their components: $\vec P = (1,\, 2m,\, m)$ $\vec Q = (4,\, -2,\, m)$
Compute the dot product: $\vec P \cdot \vec Q = (1)(4) + (2m)(-2) + (m)(m)$ $= 4 - 4m + m^2$
Since the vectors are perpendicular: $4 - 4m + m^2 = 0$ $m^2 - 4m + 4 = 0$ $(m-2)^2 = 0$
Therefore, $m = 2$
Check options:
- A: $-1$ ❌
- B: $3$ ❌
- C: $1$ ❌
- D: $2$ ✅

Hence, the correct answer is Option D.