The angle θ between the vector vecp= hati+ hatj+ hatk and unit vector along x-axis is

Question

The angle θ between the vector vecp= hati+ hatj+ hatk and unit vector along x-axis is (A) cos-1((1/√3)) (B) cos-1((1/√2)) (C) cos-1((√3/2)) (D) cos-1((1/2))

Tardigrade Admin · Accepted Answer

The angle between vecp = hat i + hat j + hat k X-axis, x = hat i is given by cos θ=( vecp ⋅ x /| vecp | x |)=(( hat i + hat j + hat k )( hat i )/√12+12+12 ⋅ √12)=(1/√3) ⇒ θ= cos -1((1/√3))

Q. The angle $θ$ between the vector $p = \hat{i} + \hat{j} + \hat{k}$ and unit vector along $x$ -axis is

$co s^{- 1} (\frac{1}{3})$

$co s^{- 1} (\frac{1}{2})$

$co s^{- 1} (\frac{3}{2})$

$co s^{- 1} (\frac{1}{2})$

The angle between
$p = \hat{i} + \hat{j} + \hat{k}$
$X$ -axis, $x = \hat{i}$
is given by
$cos θ = \frac{p \cdot x}{∣ p ∥ x ∣} = \frac{( i ^ + j ^ + k ^ ) ( i ^ )}{1 ^{2} + 1 ^{2} + 1 ^{2} \cdot 1 ^{2}} = \frac{1}{3}$
$\Rightarrow θ = cos^{- 1} (\frac{1}{3})$

Q. The angle θ between the vector p​=i^+j^​+k^ and unit vector along x-axis is

cos−1(3​1​)

cos−1(2​1​)

cos−1(23​​)

cos−1(21​)

The angle between p​=i^+j^​+k^ X-axis, x=i^ is given by cosθ=∣p​∥x∣p​⋅x​=12+12+12​⋅12​(i^+j^​+k^)(i^)​=3​1​ ⇒θ=cos−1(3​1​)

Q. The angle $θ$ between the vector $p = \hat{i} + \hat{j} + \hat{k}$ and unit vector along $x$ -axis is

$co s^{- 1} (\frac{1}{3})$

$co s^{- 1} (\frac{1}{2})$

$co s^{- 1} (\frac{3}{2})$

$co s^{- 1} (\frac{1}{2})$

The angle between
$p = \hat{i} + \hat{j} + \hat{k}$
$X$ -axis, $x = \hat{i}$
is given by
$cos θ = \frac{p \cdot x}{∣ p ∥ x ∣} = \frac{( i ^ + j ^ + k ^ ) ( i ^ )}{1 ^{2} + 1 ^{2} + 1 ^{2} \cdot 1 ^{2}} = \frac{1}{3}$
$\Rightarrow θ = cos^{- 1} (\frac{1}{3})$