Find the angle between the vectors whose direction cosines are pr… - Vidyadip

Question

Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4, 5.

✓

Answer

Let $\vec{\text{a}}$ be a vector with direction ratios 2, 3, -6.
$\Rightarrow\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 3, -4, 5.
$\Rightarrow\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}\big|\big|3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big|}$
$=\frac{6-12-30}{\sqrt{4+9+36}\sqrt{9+16+25}}$
$=\frac{-36}{\sqrt{49}\sqrt{50}}$
$=\frac{-36}{35\sqrt{2}}$
Rationalising the result, we get
$\cos\theta=-\frac{18\sqrt{2}}{35}$
$\therefore\theta=\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$
Thus, the angle between the given vectors measures $\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$.

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 "5 Marks Questions" 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

$\text{Evaluate} \lim\limits_{x \rightarrow\frac{\pi}{4}} \bigg( \frac{ \sin x - \cos x}{x- \frac{\pi}{4}} \bigg)$

Differentiate the following functions with respect to x:
$(\log\text{x})^{\cos\text{x}}$

Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}=\text{at}^2,\text{y}=2\text{at at}\text{ t}=1$

Show that $\text{A}=\begin{bmatrix} 5 & 3 \\-1 & -2 \end{bmatrix}$ satisfies the equation $x^2 - 3x - 7 = 0.$ Thus, find $A^{-1}.$

Consider the probability distribution of a random variable $X$:

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$0.1$	$0.25$	$0.3$	$0.2$	$0.15$

$\text{V}\Big(\frac{\text{X}}{2}\Big)$
Variance of $X$.

By computing the shortest distance determine whether the following pairs of lines intersect or not:3
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{-5}=\frac{\text{z}+3}{-5}$ and $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}$

Prove that: $\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2},$where $\alpha=\text{ax}-\text{by}$ and $\beta=\text{ay}+\text{bx}.$

Find the maximum and minimum value of this function.
$f(x)=\sec x+\log \cos ^2 x, 0 < x < 2 \pi $

Find the angle between the following pairs of lines:$\frac{\text{x}-2}{3}=\frac{\text{y}+3}{-2},\text{z}=5$ and $\frac{\text{x}+1}{1}=\frac{2\text{y}-3}{3}=\frac{\text{z}-5}{2}$

Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$