Home › JEE Advanced › Vectors & 3D Geometry › For vectors $\mathbf{A}=\langle 2,-1,3\rangle$ a…
For vectors $\mathbf{A}=\langle 2,-1,3\rangle$ and $\mathbf{B}=\langle 4,0,-2\rangle$, the dot product $\mathbf{A}\cdot\mathbf{B}$ is
A$2$
B$4$
C$8$
D$-2$
Answer & Solution
Correct answer: A. $2$
Using $\mathbf{A}\cdot\mathbf{B}=A_1B_1+A_2B_2+A_3B_3$, we get $2\cdot 4+(-1)\cdot 0+3\cdot(-2)=8+0-6=2$.
Related questions
Two nonzero vectors have a scalar product of zero when they are:The vector product of two vectors gives a:The dot product of two vectors returns a:Vectors that may be shifted parallel without change are called:The triangle law and the parallelogram law of addition are:Two vectors on adjacent sides of a parallelogram add along the:The relation AC = AB + BC states the:A vector equal in size but opposite in direction is the: