Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if:
A$\vec{a} \times \vec{b} = 0$
B$\vec{a} \cdot \vec{b} = 0$
C$\vec{a} + \vec{b} = 0$
D$|\vec{a}| = |\vec{b}|$
Answer & Solution
Correct answer: B. $\vec{a} \cdot \vec{b} = 0$
$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. For two non-zero vectors, this is zero exactly when $\cos\theta = 0$, i.e. $\theta = 90°$.
Option B describes *parallel* vectors (cross product zero). Option D would require the vectors to be exact opposites of each other, not just perpendicular.
Keep practising Vector Algebra
Browse all Vector Algebra questions for GRE, or explore more Mathematics topics.
Related questions
Find a unit vector in the direction of $\vec a = \hat i - 2\hat j$:Vector $\vec{AB} = 2\hat i - 4\hat j + 7\hat k$ has initial point $A(1, 5, 0)$. The terminIf vectors $2\hat i - q\hat j + 3\hat k$ and $4\hat i - 5\hat j + 6\hat k$ are collinear, The position vector of the centroid of the **tetrahedron** with vertices $A(\vec a), B(\veVectors $\vec a = \hat i - 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 6\hat j + 9\hat k$ aIf $\vec a = 4\hat i + 3\hat k$ and $\vec b = -2\hat i + \hat j + 5\hat k$, then $2\vec a The point dividing the line segment joining $A(2, -6, 8)$ and $B(-1, 3, -4)$ internally inThe distance from the point $P(2, 3, 4)$ to the **x-axis** is: