If a vector has an x component of -25.0 units and a y component of 40.0 units, then the magnitude and direction of this vector is

Question

If a vector has an x component of -25.0 units and a y component of 40.0 units, then the magnitude and direction of this vector is (A) 5√89 unit; sin-1(-5/√89) with x-axis (B) 5√89 unit; cos-1(-5/√89) with x-axis (C) 45√89 unit; cos-1(-5/9) with x-axis (D) 45√89 unit; sin-1(-5/9) with x-axis

Tardigrade Admin · Accepted Answer

| barP| = √x2 + y2 ; tan θ =(y/x) with x - axis

Q. If a vector has an x component of -25.0 units and a y component of 40.0 units, then the magnitude and direction of this vector is

$589$ unit; $sin^{- 1} \frac{- 5}{89}$ with x-axis

$589$ unit; $cos^{- 1} \frac{- 5}{89}$ with x-axis

$4589$ unit; $cos^{- 1} \frac{- 5}{9}$ with x-axis

$4589$ unit; $sin^{- 1} \frac{- 5}{9}$ with x-axis

$∣ \overset{ˉ}{P} ∣ = x^{2} + y^{2}; tan θ = \frac{y}{x}$ with $x$ - axis

Q. If a vector has an x component of -25.0 units and a y component of 40.0 units, then the magnitude and direction of this vector is

589​ unit;sin−189​−5​ with x-axis

589​ unit;cos−189​−5​ with x-axis

4589​ unit;cos−19−5​ with x-axis

4589​ unit;sin−19−5​ with x-axis

∣Pˉ∣=x2+y2​;tanθ=xy​ with x - axis

$589$ unit; $sin^{- 1} \frac{- 5}{89}$ with x-axis

$589$ unit; $cos^{- 1} \frac{- 5}{89}$ with x-axis

$4589$ unit; $cos^{- 1} \frac{- 5}{9}$ with x-axis

$4589$ unit; $sin^{- 1} \frac{- 5}{9}$ with x-axis

$∣ \overset{ˉ}{P} ∣ = x^{2} + y^{2}; tan θ = \frac{y}{x}$ with $x$ - axis