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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Consider the following diagram, where the forces in the cable are given. Based on the above information, answer the following questions.
  1. The equation of line along the cable AD is:
  1. $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{\text{z}-30}{15}$
  2. $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{\text{z}-30}{15}$
  3. $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{30-\text{z}}{15}$
  4. $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
  1. The length of cable DC is:
  1. $4\sqrt{61}\text{m}$
  2. $5\sqrt{61}\text{m}$
  3. $6\sqrt{61}\text{m}$
  4. $7\sqrt{61}\text{m}$
  1. The vector DB is:
  1. $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
  2. $6\hat{\text{i}}-4\hat{\text{j}}+30\hat{\text{k}}$
  3. $6\hat{\text{i}}+4\hat{\text{j}}+30\hat{\text{k}}$
  4. None of these
  1. The sum of vectors along the cables is:
  1. $17\hat{\text{i}}+6\hat{\text{j}}+90\hat{\text{k}}$
  2. $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
  3. $17\hat{\text{i}}+6\hat{\text{j}}-90\hat{\text{k}}$
  4. None of these
  1. The sum of distances of points A, B and C from the origin, i.e., OA + OB + OC is:
  1. $\sqrt{164}+\sqrt{52}+\sqrt{625}$
  2. $\sqrt{52}+\sqrt{625}+\sqrt{48}$
  3. $\sqrt{164}+\sqrt{625}+\sqrt{49}$
  4. None of these
✓

Answer

  1. (d) $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
Solution:

Clearly, the coordinates of A are (8, 10, 0) and D are (0, 0, 30)

$\therefore$ Equation of AD is given by

$\frac{\text{x}-0}{8-0}=\frac{\text{y}-0}{10-0}=\frac{30-\text{z}}{-30}$

$\Rightarrow\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
  1. (b) $5\sqrt{61}\text{m}$
Solution:

The coordinates of point Care (15, -20, 0) and D are (0, 0, 30)

$\therefore$ Length of the cable DC

$=\sqrt{(0-15)^2+(0-20)^2+(30-0)^2}$

$=\sqrt{225+400+900}$

$=\sqrt{1525}=\sqrt{61}\text{m}.$
  1. (a) $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
Solution:

Since, the coordinates of point B are (-6, 4, 0) and D are (0, 0, 30), therefore vector DB is

$(-6-0)\hat{\text{i}}+(4-0)\hat{\text{j}}+(0-30)\hat{\text{k}},$

i.e., $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
  1. (b) $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
​​​​​​​Solution:

Required sum

$(8\hat{\text{i}}+10\hat{\text{j}}-30\hat{\text{k}})+(-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}})+(15\hat{\text{i}}-20\hat{\text{j}}-30\hat{\text{k}})$

$17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
  1. (a) $\sqrt{164}+\sqrt{52}+\sqrt{625}$
​​​​​​​Solution:

Clearly, $\text{OA}=\sqrt{8^2+10^2}=\sqrt{164}$

$\text{OB}=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$

And $\text{OC}=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}$

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