Q108 / 180 · KCET 2024 · Mathematics

The vectors $\mathbf{A B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}$ and $\mathbf{A C}=5 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ are the sides of a $\triangle A B C$, The length of the median through $A$ is

  • A
    $\sqrt{18}$
  • B
    $\sqrt{72}$
  • Correct answer
    C
    $\sqrt{33}$
  • D
    $\sqrt{288}$

Solution

Let $\mathbf{A}=$ origin $\therefore \mathbf{A B}=$ Position vector of $\mathbf{B}$ $\mathbf{A C}=$ Position vector of $\mathbf{C}$ $\therefore$ Position vector of mid-point of $\mathbf{B}$ and $\mathbf{C}$ $=\mathbf{P}=\frac{\mathbf{A B}+\mathbf{A C}}{2}$ $=\left(\frac{3+5}{2}\right) \hat{\mathbf{i}}+\left(\frac{0-2}{2}\right) \hat{\mathbf{j}}+\left(\frac{4+4}{2}\right) \hat{\mathbf{k}}$ $=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ Median $=\mathbf{A P}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ Hence, length of median $=|\mathbf{A P}|=\sqrt{16+1+16}$ $=\sqrt{33}$

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

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