Download App

Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry3 Marks
Question
Find the angle between the lines whose direction ratios are a, b, c and b - c, c - a, a - b.
✓

Answer

Direction ratios of one line are a, b, c 
$\Rightarrow $ A vector along this line is ${\vec b_1} = a\hat i + b\hat j + c\hat k$ 
Direction ratios of second line are $b - c,c - a,a - b$ 
$\Rightarrow$ A vector along second line is ${\vec b_2} = \left( {b - c} \right)\hat i + \left( {c - a} \right)\hat j + \left( {a - b} \right)\hat k$
Let $\theta $ be the angle between the two lines, then
$\cos \theta = \frac{{\left| {{{\vec b}_1}.{{\vec b}_2}} \right|}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}} = \frac{{a\left( {b - c} \right) + b\left( {c - a} \right) + c\left( {a - b} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }}$
$= \frac{{ab - ac + bc - ab + ac - bc}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }} = 0 = \cos {90^0}$
$\Rightarrow \theta = {90^0}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks Question" from Three Dimensional Geometry→Three Dimensional Geometry — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
→
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
→
Evalute the following integrals:
$\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
→
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}\text{x}-1&1&1\\1&\text{x}-1&1\\1&1&\text{x}-1 \end{vmatrix}$
→
Write the projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}}$ when $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$
→
Find the value of $\tan^{-1}\Big(-\frac{1}{\sqrt{3}}\Big)+\cot^{-1}\Big(\frac{1}{\sqrt{3}}\Big)+\tan^{-1}\Big[\sin\Big(\frac{-\pi}{2}\Big)\Big].$
→
Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y\ ' = e^x \sin x.$
→
$\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}, \overrightarrow{b} = 3\hat{i} - \hat{j} + 2\hat{k}, \text{show that}\bigg(\overrightarrow{a} +\overrightarrow{b}\bigg) \text{and} \bigg(\overrightarrow{a} -\overrightarrow{b}\bigg)$ are perpendicular to each other.
→
Solve for x:
$\tan^{-1} \bigg(\frac{x - 2}{x - 1}\bigg) + \tan^{-1} \bigg(\frac{x + 2}{x + 1}\bigg) = \frac{\pi}{4}$
→
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$
→