alleryThe triangle inequality for vectors states:
A$|\vec a + \vec b| = |\vec a| + |\vec b|$
B$|\vec a + \vec b| \le |\vec a| + |\vec b|$
C$|\vec a + \vec b| \ge |\vec a| + |\vec b|$
D$|\vec a + \vec b| = |\vec a| - |\vec b|$
Answer & Solution
Correct answer: B. $|\vec a + \vec b| \le |\vec a| + |\vec b|$
In any triangle, the third side cannot exceed the sum of the other two: $|\vec a + \vec b| \le |\vec a| + |\vec b|$, with equality only when $\vec a$ and $\vec b$ are parallel and same-direction.
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: