Advertisements
Advertisements
प्रश्न
If the lines `(x - 1)/2 = (y + 1)/3 = (z - 1)/4 and (x - 3)/1 = (y - k)/2 = z/1` intersect each other, then find k.
Advertisements
उत्तर
The lines `(x - x_1)/l_1 = (y - y_1)/m_1 = (z - z_1)/n_1 and (x - x_2)/(l_2) = (y - y_2)/m_2 = (z - z_2)/n_2`
intersect, if `|(x_2 - x_1, y_2 - y_1,z_2 - z_1),(l_1, m_1, n_1),(l_2, m_2, n_2)|` = 0
The equations of the given lines are
`(x - 1)/(2) = (y +1)/3 = (z - 1)/4 and (x - 3)/1 = (y - k)/2 = z/1`
∴ x1 = 1, y1 = –1, z1 = 1, x2 = 3, y2 = k, z2 = 0,
l1 = 2, m1 = 3, n1 = 4, l2 = 1, m2 = 2, n2 = 1.
Since these lines intersect, we get
`|(2, k + 1, -1),(2, 3, 4),(1, 2, 1)|` = 0
∴ 2(3 – 8) – (k + 1)(2 – 4) – 1(4 – 3) = 0
∴ – 10 + 2(k + 1) – 1 = 0
∴ 2(k + 1) = 11
∴ k + 1 `11/2`
∴ k = `9/2`
APPEARS IN
संबंधित प्रश्न
Find the length of the perpendicular (2, –3, 1) to the line `(x + 1)/(2) = (y - 3)/(3) = (z + 1)/(-1)`.
Find the co-ordinates of the foot of the perpendicular drawn from the point `2hati - hatj + 5hatk` to the line `barr = (11hati - 2hatj - 8hatk) + λ(10hati - 4hatj - 11hatk).` Also find the length of the perpendicular.
A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector `2hati + hatj - 2hatk`.
Find the vector equation of the plane passing through the point having position vector `hati + hatj + hatk` and perpendicular to the vector `4hati + 5hatj + 6hatk`.
Show that the line `bar"r" = (2hat"j" - 3hat"k") + lambda(hat"i" + 2hat"j" + 3hat"k") and bar"r" = (2hat"i" + 6hat"j" + 3hat"k") + mu(2hat"i" + 3hat"j" + 4hat"k")` are coplanar. Find the equation of the plane determined by them.
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line `(x + 3)/(5) = (y - 1)/(2) = (z + 4)/(3)`.
Choose correct alternatives :
Equation of X-axis is ______.
The perpendicular distance of the plane 2x + 3y – z = k from the origin is `sqrt(14)` units, the value of k is ______.
Choose correct alternatives :
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
Choose correct alternatives :
The direction cosines of the normal to the plane 2x – y + 2z = 3 are ______
Choose correct alternatives :
The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is
Solve the following :
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z - 7 = 0
Solve the following :
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
The equation of X axis is ______
If the foot of the perpendicular drawn from the origin to the plane is (4, −2, -5), then the equation of the plane is ______
The coordinates of the foot of perpendicular drawn from the origin to the plane 2x + y − 2z = 18 are ______
If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation
Find the perpendicular distance of origin from the plane 6x − 2y + 3z - 7 = 0
Show that the lines `(x + 1)/(-10) = (y + 3)/(-1) = (z - 4)/(1)` and `(x + 10)/(-1) = (y + 1)/(-3) = (z - 1)/4` intersect each other.also find the coordinates of the point of intersection
If z1 and z2 are z-coordinates of the points of trisection of the segment joining the points A (2, 1, 4), B (–1, 3, 6) then z1 + z2 = ______.
The equation of a plane containing the point (1, - 1, 2) and perpendicular to the planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8 is ______.
If the line `(x - 3)/2 = (y + 2)/-1 = (z + 4)/3` lies in the plane lx + my - z = 9, then l2 + m2 is equal to ______
If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is ______
The equation of the plane passing through the point (– 1, 2, 1) and perpendicular to the line joining the points (– 3, 1, 2) and (2, 3, 4) is ______.
XY-plane divides the line joining the points A(2, 3, -5) and B(1, -2, -3) in the ratio ______
The equation of the plane through the point (2, -1, -3) and parallel to the lines `(x - 1)/3 = (y + 2)/2 = z/(-4)` and `x/2 = (y - 1)/(-3) = (z - 2)/2` is ______
If the plane passing through the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) is ax + by + cz = d then a + 2b + 3c = ______.
The equation of the plane passing through a point having position vector`-2hat"i" + 7hat"j" + 5hat"k"` and parallel to the vectors `4hat"i" - hat"j" + 3hat"k"` and `hat"i" + hat"j" + hat"k"` is ______.
If the mirror image of the point (2, 4, 7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to ______.
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line `vecr = -hatk + λ(hati + hatj + 2hatk)`, λ ∈ R. Then, which of the following points lies on T?
Let P be a plane passing through the points (1, 0, 1), (1, –2, 1) and (0, 1, –2). Let a vector `vec"a" = αhat"i" + βhat"j" + γhat"k"` be such that `veca` is parallel to the plane P, perpendicular to `(hat"i"+2hat"j"+3hat"k")`and `vec"a".(hat"i" + hat"j" + 2hat"j")` = 2, then (α – β + γ)2 equals ______.
The equation of the plane through the line x + y + z + 3 = 0 = 2x – y + 3z + 1 and parallel to the line `x/1 = y/2 = z/3`, is ______.
The equation of the plane passes through the point (2, 5, –3) perpendicular to the plane x + 2y + 2z = 1 and x – 2y + 3z = 4 is ______.
What will be the equation of plane passing through a point (1, 4, – 2) and parallel to the given plane – 2x + y – 3z = 9?
If the foot of the perpendicular drawn from the origin to the plane is (4, –2, 5), then the equation of the plane is ______.
Find the vector equation of the line passing through the point (–2, 1, 4) and perpendicular to the plane `barr*(4hati - 5hatj + 7hatk)` = 15
Find the point of intersection of the line `(x + 1)/2 = (y - 1)/3 = (z - 2)/1` with the plane x + 2y – z = 6.
Find the equation of the plane containing the line `x/(-2) = (y - 1)/3 = (1 - z)/1` and the point (–1, 0, 2).
