If $A(1,1,2), B(4,2,1)$ and $C(2,3,5)$ are the vertices of a triangle, then a vector representing the median of the triangle through $A$ is

• A

  $3 \hat{i}+4 \hat{j}+5 \hat{k}$
• Correct answer
  B

  $(1+2 t) \hat{\mathrm{i}}+\left(1+\frac{3 t}{2}\right) \hat{\mathrm{j}}+(2+t) \hat{\mathrm{k}}$
• C

  $2 t \hat{\mathrm{i}}+(7 t-1) \hat{\mathrm{j}}+5 t^2 \hat{\mathrm{k}}$
• D

  $7 t^2 \hat{i}+6 \hat{j}+4 \hat{k}$

Solution

Since $A D$ is median $\therefore \quad D=\left(\frac{4+2}{2}, \frac{2+3}{2}, \frac{1+5}{2}\right)=\left(3, \frac{5}{2}, 3\right)$ $\therefore$ Equation of $A D$ is $\begin{aligned} \mathbf{r} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t\left((3-1) \hat{\mathbf{i}}+\left(\frac{5}{2}-1\right) \hat{\mathbf{j}}+(3-2) \hat{\mathbf{k}}\right) \\ & =\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+t\left(2 \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}+\hat{\mathbf{k}}\right) \\ & =(1+2 t) \hat{\mathbf{i}}+\left(1+\frac{3}{2} t\right) \hat{\mathbf{j}}+(2+t) \hat{\mathbf{k}}\end{aligned}$

Hence, the correct answer is option: 'B'.