alleryThe projection of vector $\vec A$ on vector $\vec B$ is given by
A$\dfrac{\vec A\times\vec B}{|\vec B|}$
B$\dfrac{\vec A\cdot\vec B}{|\vec B|}$
C$\dfrac{|\vec A|}{\vec B}$
D$\dfrac{\vec A\cdot\vec B}{|\vec A|}$
Answer & Solution
Correct answer: B. $\dfrac{\vec A\cdot\vec B}{|\vec B|}$
The notes explicitly give the projection of $\vec A$ on $\vec B$ as $\dfrac{\vec A\cdot\vec B}{|\vec B|}$.
This is the scalar projection along $\vec B$.
Keep practising Unit 8A: Areas Between Curves
Browse all Unit 8A: Areas Between Curves questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
Let $\mathbf{A}=\langle 1,2,3\rangle$, $\mathbf{B}=\langle 2,0,1\rangle$, and $\mathbf{C}=For nonzero vector $\mathbf{A}$, which expression gives a vector in the same direction as When subtracting vectors, $\vec A-\vec B$ is evaluated asWhich statement about a null vector is correct?If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.Find the modulus of the complex number $z = 3 + 4i$.The locus of the point $z$ in the Argand plane satisfying $|z - 2| = |z + 2|$ is:Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.