Question

Q.The two vectors j+k and 3i-j+4k represents the two coiniatial side of triangle ABC.Find the length of median through coinitial point A.

Answer 1

Hope this will be useful to you.

Answer 2

Answer:

Median’s length through coinitial point A is $\sqrt{\frac{17}{2}}$ units.

To find:

Median’s length through coinitial point A

Solution:

From the diagram,  

In ∆ ABC, let AD be the median to side BC.

$\overline { A B } = 0 i + j + k \text { and } \overline { A C } = 3 i - j + 4 k$

Since  D is the midpoint of BC,  

$B D : D C = 1 : 1$

Thus, using Section Formula,

$\begin{array} { c } { \overline { A D } = \frac { m _ { 1 } ( \overline { A B } ) + m _ { 2 } ( \overline { A C } ) } { m _ { 1 } + m _ { 2 } } } \\\\ { = \frac { 1 ( 0 i + j + k ) + 1 ( 3 i - j + 4 k ) } { 1 + 1 } } \\\\ { = \frac { j + k + 3 i - j + 4 k } { 1 + 1 } } \\\\ { = \frac { 3 i + 0 j + 5 k } { 2 } } \\\\ { = \frac { 3 } { 2 } i + 0 j + \frac { 5 } { 2 } k } \\\\ { = \frac { 3 } { 2 } i + 0 j + \frac { 5 } { 2 } k } \end{array}$

$| \overline { A D } | = \sqrt { \left( \frac { 3 } { 2 } \right) ^ { 2 } + ( 0 ) ^ { 2 } + \left( \frac { 5 } { 2 } \right) ^ { 2 } } u n i t s$$\begin{array} { l } { = \sqrt { \frac { 9 } { 4 } + 0 + \frac { 25 } { 4 } } \text { units } } \\\\ { = \sqrt { \frac { 9 + 25 } { 4 } } \text { units } } \\\\ { = \sqrt { \frac { 34 } { 4 } } = \sqrt { \frac { 17 } { 2 } } \text { units. } } \end{array}$

Median’s length through coinitial point A is $\sqrt{\frac{17}{2}}$ units.