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Algebra of Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
The two vectors $\hat{\text{j}}+\hat{\text{k}}$ and $3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ represents the sides $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{AC}}$ respectively of a triangle ABC. Find the length of the median through A.
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Answer

Disclaimer: The question has been solved by taking the vector $\overrightarrow{\text{AB}}$ as $\hat{\text{j}}+\hat{\text{k}}$.In $\triangle\text{ABC},\ \overrightarrow{\text{AB}}=\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{AC}}=3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
Let the position vector of A be (0, 0, 0). Then, the position vectors of B and C are (0, 1, 1) and (3, -1, 4), respectively.

Suppose D be the mid-point of the line segment joining the points B(0, 1, 1) and C(3, -1, 4).
$\therefore$ position vector of D $=\frac{\big(\hat{\text{j}}+\hat{\text{k}}\big)+\big(3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)}{2}=\frac{3\hat{\text{i}}+5\hat{\text{k}}}{2}=\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}$
Now,
Length of the median, AD =
$\Big|\overrightarrow{\text{AD}}\Big|=\Big|\Big(\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}\Big)-\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)\Big|$
$=\Big|\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}\Big|$
$=\sqrt{\Big(\frac{3}{2}\Big)^2+0^2+\Big(\frac{5}2\Big)^2}$
$=\sqrt{\frac{34}{4}}$
$=\sqrt{\frac{17}2}\text{units}$

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