← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Scalar and Vector Fields · Vector Analysis · Read the full method →

Question

Prove that the vectors $\vec a=3\hat i+\hat j-2\hat k,\ \vec b=-\hat i+3\hat j+4\hat k,\ \vec c=4\hat i-2\hat j-6\hat k$ can form the sides of a triangle. Find the lengths of the medians of the triangle.

Technique

Triangle test = one side is the sum of the other two ($\vec a=\vec b+\vec c$); place vertices, take midpoints, compute median vectors and norms.

Solution

Step 1 — Show the three vectors close into a triangle

Three vectors can be the sides of a triangle iff one is the (vector) sum of the other two, equivalently their directed sides close head-to-tail. Compute

Hence $\vec a=\vec b+\vec c$, i.e. $\vec b+\vec c-\vec a=\vec 0$. Placing the sides head-to-tail ($\vec b$ then $\vec c$) returns to the start of $\vec a$, so the three vectors form a closed triangle. $\blacksquare$

Step 2 — Fix the vertices

Take vertex $A$ at the origin, $B=A+\vec b$, $C=B+\vec c$. Then

The sides are $\overrightarrow{AB}=\vec b,\ \overrightarrow{BC}=\vec c,\ \overrightarrow{AC}=\vec a$, with lengths

Step 3 — Median from $A$ (to midpoint of $BC$)

Midpoint of $BC$: $M_A=\dfrac{B+C}{2}=\left(1,2,1\right)$. Median vector $\overrightarrow{AM_A}=(1,2,1)$, length

Step 4 — Median from $B$ (to midpoint of $AC$)

$M_B=\dfrac{A+C}{2}=\left(\tfrac32,\tfrac12,-1\right)$, $\overrightarrow{BM_B}=\left(\tfrac52,-\tfrac52,-5\right)$,

Step 5 — Median from $C$ (to midpoint of $AB$)

$M_C=\dfrac{A+B}{2}=\left(-\tfrac12,\tfrac32,2\right)$, $\overrightarrow{CM_C}=\left(-\tfrac72,\tfrac12,4\right)$,

Answer