Advertisements
Advertisements
प्रश्न
The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is ______.
विकल्प
0°
30°
45°
90°
Advertisements
उत्तर
The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is 90°.
Explanation:
2x = 3y = – z
6x = – y = – 4z
`x/(1/2) = y/(1/3) = z/-1`
`x/(1/6) = y/-1 = z/((-1)/4)`
Direction cosine's of both lines are `(1/2, 1/3, -1)` and `(1/6, -1, -1/4)` respectively.
(l1, m1, n1) = (3, 2, – 6) and (l2, m2, n2) = (2, – 12, – 3)
cos θ = l1l2 + m1m2 + n1n2
= 6 – 24 + 18
= 0
`\implies` cos θ = 0 = cos 90°
∴ θ = 90°.
APPEARS IN
संबंधित प्रश्न
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-2)/2 = (y-1)/5 = (z+3)/(-3)` and `(x+2)/(-1) = (y-4)/8 = (z -5)/4`
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 values of p so the line `(1-x)/3 = (7y-14)/2p = (z-3)/2` and `(7-7x)/(3p) = (y -5)/1 = (6-z)/5` are at right angles.
Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.
Find the angle between the line \[\vec{r} = \left( 2 \hat{i}+ 3 \hat {j} + 9 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] and the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 5 .\]
Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3x − y + z = 1.
The line \[\vec{r} = \hat{i} + \lambda\left( 2 \hat{i} - m \hat{j} - 3 \hat{k} \right)\] is parallel to the plane \[\vec{r} \cdot \left( m \hat{i} + 3 \hat{j} + \hat{k} \right) = 4 .\] Find m.
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.
Find the angle between the line
Write the angle between the line \[\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{- 2}\] and the plane x + y + 4 = 0.
Find the angle between the two lines `2x = 3y = -z and 6x =-y = -4z`
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
`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 ______.
Find the angle between the following two lines:
`vecr = 2hati - 5hatj + hatk + λ(3hati + 2hatj + 6hatk)`
`vecr = 7hati - 6hatk + μ(hati + 2hatj + 2hatk)`
For direction ratios \[(a_1,b_1,c_1)\] and \[(a_2,b_2,c_2)\], which expression gives \[\cos\theta\]?
If \[(l_1,m_1,n_1)\] and \[(l_2,m_2,n_2)\] are direction cosines of two lines, then \[\cos\theta\] equals:
Two lines with direction ratios \[(a_1,b_1,c_1)\] and \[(a_2,b_2,c_2)\] are perpendicular when:
Two lines with direction ratios \[(a_1,b_1,c_1)\] and \[(a_2,b_2,c_2)\] are parallel when:
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?
