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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry1 Mark
MCQ
The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:
  • A
    45°
  • B
    30°
  • C
    60°
  • ✓
    90°
✓

Answer

Correct option: D.
90°
We have

$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$

$\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$

The direction of the given lines are propotional to 2, 5, 4 are 1, 2, -3.

The given lines are parallel to the vectors $\vec{\text{b}}_1=2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}.$

Let $\theta$ be the angle between the given lines.

Now,

$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$

$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$

$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$

$\Rightarrow\theta=90^\circ$

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