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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA4 Marks
Question
A building is to be constructed in the form of a triangular pyramid, ABCD as shown in the figure.

Let its angular points are A(0, 1, 2), B(3, 0, 1), C(4, 3, 6), and D(2, 3, 2), and G be the point of intersection of the medians of $\triangle\text{BCD}.$
Based on the above information, answer the following questions.
  1. The coordinates of point Gare:
  1. (2, 3, 3)
  2. (3, 3, 2)
  3. (3, 2, 3)
  4. (0, 2, 3)
  1. The length of vector $\overline{\text{AG}}$ is:
  1. $\sqrt{17}\text{ units}$
  2. $\sqrt{11}\text{ units}$
  3. $\sqrt{13}\text{ units}$
  4. $\sqrt{19}\text{ units}$
  1. Area of $\triangle\text{ABC}$ (in sq. units) is:
  1. $\sqrt{10}$
  2. $2\sqrt{10}$
  3. $3\sqrt{10}$
  4. $5\sqrt{10}$
  1. The sum of lengths of $\overline{\text{AB}}$ and $\overline{\text{AC}}$ is:
  1. 5 units
  2. 9.32 units
  3. 10 units
  4. 11 units
  1. The length of the perpendicular from the vertex D on the opposite face is:
  1. $\frac{6}{\sqrt{10}}\text{ units}$
  2. $\frac{2}{\sqrt{10}}\text{ units}$
  3. $\frac{3}{\sqrt{10}}\text{ units}$
  4. $8\sqrt{10}\text{ units}$
✓

Answer

  1. (c) (3, 2, 3)
Solution:
Clearly, G be the centroid of $\triangle\text{BCD},$ therefore coordinates of G are
$\Big(\frac{3+4+2}{3},\frac{0+3+3}{3},\frac{1+6+2}{3}\Big)$
= (3, 2, 3)
  1. (b) $\sqrt{11}\text{ units}$
​​​​​​​​​​​​​​Solution:
Since, $\text{A}\equiv(0,1,2)$ and G = (3, 2, 3)
$\therefore\overline{\text{AG}}=(3-0)\hat{\text{i}}+(2-1)\hat{\text{j}}+(3-2)\hat{\text{k}}=3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow|\overline{\text{AG}}|^2=3^2+1^2+1^2=9+1+1=11$
$\Rightarrow|\overline{\text{AG}}|=\sqrt{11}$
  1. (c) $3\sqrt{10}$
​​​​​​​​​​​​​​​​​​​​​Solution:
Clearly, area of $\triangle\text{ABC}=\frac{1}{2}|\overline{\text{AB}}\times\overline{\text{AC}}|$
Here, $\overline{\text{AB}}\times\overline{\text{AC}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3-0&0-1&1-2\\4-0&3-1&6-2\end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&-1\\4&2&4\end{vmatrix}$
$=\hat{\text{i}}(-4+2)-\hat{\text{j}}(12+4)+\hat{\text{k}}(6+4)$
$=-2\hat{\text{i}}-16\hat{\text{j}}+10\hat{\text{k}}$
$\therefore|\overline{\text{AB}}\times\overline{\text{AC}}|=\sqrt{(-2)^2+(-16)^2+10^2}$
$=\sqrt{4+256+100}=\sqrt{360}=6\sqrt{10}$
Hence, area of  $\triangle\text{ABC}=\frac{1}{2}\times6\sqrt{10}=3\sqrt{10}\text{sq}.\text{units}.$
  1. (b) 9.32 units
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:
Here,​​​​​​​ $\overline{\text{AB}}=3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\Rightarrow|\overline{\text{AB}}|=\sqrt{9+1+1}=\sqrt{11}$
Also, $\overline{\text{AC}}=4\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\Rightarrow|\overline{\text{AC}}|=\sqrt{16+4+16}=\sqrt{36}=6$
Now, $|\overline{\text{AB}}|+\overline{\text{AC}}|=\sqrt{11}+6$
= 3.32 + 6 = 9.32 units.
  1. (a) $\frac{6}{\sqrt{10}}\text{ units}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:
The length of the perpendicular from the vertex D on the opposite face
$=|\text{Projection of }\overline{\text{AC}}\text{ on}\overline{\text{ AB}}\times\overline{\text{AC}}|$
$=\begin{vmatrix}\frac{(2\hat{\text{i}}+2\hat{\text{j}})\cdot(-2\hat{\text{i}}+16\hat{\text{j}}+10\hat{\text{k}})}{\sqrt{(-2)^2+(-6)^2+10^2}}\\\end{vmatrix}$
$=\begin{vmatrix}\frac{-4-32}{\sqrt{360}}\\\end{vmatrix}=\frac{36}{6\sqrt{10}}$
$=\frac{6}{\sqrt{10}}\text{ units}.$

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