Q107 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
The coordinates of the vertices of the triangle are $A(-2,3,6), B(-4,4,9)$ and $C(0,5,8)$. The direction cosines of the median $\mathrm{BE}$ are
- A$\left\langle-\dfrac{3}{\sqrt{13}}, \quad 0, \quad-\dfrac{2}{\sqrt{13}}\right\rangle$
- Correct answerB$\left\langle\dfrac{3}{\sqrt{13}}, \quad 0, \quad-\dfrac{2}{\sqrt{13}}\right\rangle$
- C$\left\langle\dfrac{3}{4}, \quad 0, \quad-\dfrac{2}{4}\right\rangle$
- D$\left\langle 1, \quad 0, \quad-\dfrac{2}{3}\right\rangle$
Solution
The median BE connects vertex B to the midpoint E of side AC. The coordinates of A are $(-2, 3, 6)$ and C are $(0, 5, 8)$.
The midpoint E of AC is given by $E = \left( \dfrac{-2 + 0}{2}, \dfrac{3 + 5}{2}, \dfrac{6 + 8}{2} \right) = (-1, 4, 7)$.
The vector $\overrightarrow{BE}$ is given by $E - B = (-1 - (-4), 4 - 4, 7 - 9) = (3, 0, -2)$.
The magnitude of vector $\overrightarrow{BE}$ is $|\overrightarrow{BE}| = \sqrt{3^2 + 0^2 + (-2)^2} = \sqrt{9 + 0 + 4} = \sqrt{13}$.
The direction cosines of the vector $\overrightarrow{BE}$ are obtained by dividing the components of the vector by its magnitude:
$\left( \dfrac{3}{\sqrt{13}}, \dfrac{0}{\sqrt{13}}, \dfrac{-2}{\sqrt{13}} \right) = \left( \dfrac{3}{\sqrt{13}}, 0, -\dfrac{2}{\sqrt{13}} \right)$.
Answer: $\left\langle\dfrac{3}{\sqrt{13}}, \quad 0, \quad-\dfrac{2}{\sqrt{13}}\right\rangle$
The midpoint E of AC is given by $E = \left( \dfrac{-2 + 0}{2}, \dfrac{3 + 5}{2}, \dfrac{6 + 8}{2} \right) = (-1, 4, 7)$.
The vector $\overrightarrow{BE}$ is given by $E - B = (-1 - (-4), 4 - 4, 7 - 9) = (3, 0, -2)$.
The magnitude of vector $\overrightarrow{BE}$ is $|\overrightarrow{BE}| = \sqrt{3^2 + 0^2 + (-2)^2} = \sqrt{9 + 0 + 4} = \sqrt{13}$.
The direction cosines of the vector $\overrightarrow{BE}$ are obtained by dividing the components of the vector by its magnitude:
$\left( \dfrac{3}{\sqrt{13}}, \dfrac{0}{\sqrt{13}}, \dfrac{-2}{\sqrt{13}} \right) = \left( \dfrac{3}{\sqrt{13}}, 0, -\dfrac{2}{\sqrt{13}} \right)$.
Answer: $\left\langle\dfrac{3}{\sqrt{13}}, \quad 0, \quad-\dfrac{2}{\sqrt{13}}\right\rangle$
Hence, the correct answer is option: 'B'.