Advertisements
Advertisements
Question
Find the angle between the following two lines:
`vecr = 2hati - 5hatj + hatk + λ(3hati + 2hatj + 6hatk)`
`vecr = 7hati - 6hatk + μ(hati + 2hatj + 2hatk)`
Advertisements
Solution
Angle between two vectors
`vecr = veca_1 + λvecb_1` and `vecr = veca_2 + μvecb_2` is given by
cos θ = `|(vecb_1 . vecb_2)/(|vecb_1||vecb_2|)|`
`vecr = (2hati - 5hatj + hatk) + λ(3hati + 2hatj + 6hatk)`
`veca_1 = 2hati - 5hatj + hatk, vecb_1 = 3hati + 2hatj + 6hatk`
`vecr = (7hati - 6hatk) + μ(hati + 2hatj + 2hatk)`
`veca_2 = (7hati - 6hatk), vecb_2 = hati + 2hatj + 2hatk`
Now `vecb_1 . vecb_2 = (3hati + 2hatj + 6hatk) . (hati + 2hatj + 2hatk)`
= 3 + 4 + 12
= 19
`|vecb_1| = sqrt(3^2 + 2^2 + 6^2)`
= `sqrt(49)`
= 7
`|vecb_2| = sqrt(1^2 + 2^2 + 2^2)`
= `sqrt(9)`
= 3
cos θ = `|b_1 . b_2|/(|b_1||b_2|)`
cos θ = `|19/(7 xx 3)|`
= `19/21`
θ = `cos^-1 19/21`
APPEARS IN
RELATED QUESTIONS
If the angle between the lines represented by ax2 + 2hxy + by2 = 0 is equal to the angle between the lines 2x2 - 5xy + 3y2 =0,
then show that 100(h2 - ab) = (a + b)2
Find the angle between the following pairs of lines:
`x/y = y/2 = z/1` and `(x-5)/4 = (y-2)/1 = (z - 3)/8`
Find the angle between the line \[\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{1}\] and the plane 2x + y − z = 4.
Show that the line whose vector equation is \[\vec{r} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}+ \lambda\left( \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] is parallel to the plane whose vector \[\vec{r} \cdot \left( \hat{i} + \hat{j} - \hat{k} \right) = 7 .\] Also, find the distance between them.
Find the angle between the line \[\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}\] and the plane
3x + 4y + z + 5 = 0.
Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 1\] and the line whose vector equation is \[\vec{r} = \left( - \hat{i} + \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right)\] are parallel. Also, find the distance between them.
Find the angle between the line
Find the angle between the lines whose direction cosines are given by the equations: 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0.
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0.
`vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)` and `vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)`
Assertion (A): The acute angle between the line `barr = hati + hatj + 2hatk + λ(hati - hatj)` and the x-axis is `π/4`
Reason(R): The acute angle 𝜃 between the lines `barr = x_1hati + y_1hatj + z_1hatk + λ(a_1hati + b_1hatj + c_1hatk)` and `barr = x_2hati + y_2hatj + z_2hatk + μ(a_2hati + b_2hatj + c_2hatk)` is given by cosθ = `(|a_1a_2 + b_1b_2 + c_1c_2|)/sqrt(a_1^2 + b_1^2 + c_1^2 sqrt(a_2^2 + b_2^2 + c_2^2)`
The angle between two lines `(x + 1)/2 = (y + 3)/2 = (z - 4)/(-1)` and `(x - 4)/1 = (y + 4)/2 = (z + 1)/2` is ______.
A straight line L through the point (3, –2) is inclined at an angle of 60° to the line `sqrt(3)x + y` = 1. If L also intersects the x-axis, then the equation of L is ______.
The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is ______.
If \[(l_1,m_1,n_1)\] and \[(l_2,m_2,n_2)\] are direction cosines of two lines, then \[\cos\theta\] equals:
For direction ratios \[(a_1,b_1,c_1)\] and \[(a_2,b_2,c_2)\], which expression gives \[\sin\theta\]?
Two lines with direction ratios \[(a_1,b_1,c_1)\] and \[(a_2,b_2,c_2)\] are perpendicular when:
For the lines \[\vec r=\vec a_1+\lambda\vec b_1\] and \[\vec r=\vec a_2+\mu\vec b_2\], the angle between the lines is the angle between:
In the symmetric or Cartesian form \[\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\], the direction ratios are:
For direction vectors \[\vec b_1=(1,2,2)\] and \[\vec b_2=(3,2,6)\], the angle between the lines is:
If two lines do not pass through the origin, what should be imagined to determine the angle between them?
What is generally the required angle between two lines?
