Advertisements
Advertisements
प्रश्न
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (– 2, 3, 4) and parallel to the straight line `(x - 1)/(-4) = (y + 3)/5 = (8 - z)/6`
उत्तर
`vec"a" = -2hat"i" + 3hat"j" + 4hat"k" (x_1 y_1, z_1)` = (– 2, 3, 4)
`vec"b" = - 4hat"i" + 5hat"j" + 6hat"k" (l, "m", "n")` = (– 4, 5, 6)
Vector equation
`vec"r" = vec"a" + "t"vec"b"`, t ∈ R
`vec"r" = (-2hat"i" + 3hat"j" + 4hat"k") + "t"(-4hat"i" + 5hat"j" - 6hat"k")`, t ∈ R
Cartessain equation
`(x - x_1)/1 = (y - y_1)/"m" = (z - z_1)/"n"`
`(x + 2)/(-4) = (y - 3)/5 = (z - 4)/(- 6)`
APPEARS IN
संबंधित प्रश्न
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector `4hat"i" + 3hat"j" - 7hat"k"` and parallel to the vector `2hat"i" - 6hat"j" + 7hat"k"`
Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7, 9, 13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points
Find the acute angle between the following lines.
`vec"r" = (4hat"i" - hat"j") + "t"(hat"i" + 2hat"j" - 2hat"k")`
Find the acute angle between the following lines.
`(x + 4)/3 = (y - 7)/4 = (z + 5)/5, vec"r" = 4hat"k" + "t"(2hat"i" + hat"j" + hat"k")`
Find the acute angle between the following lines.
2x = 3y = – z and 6x = – y = – 4z
The vertices of ΔABC are A(7, 2, 1), 5(6, 0, 3), and C(4, 2, 4). Find ∠ABC
f the straight line joining the points (2, 1, 4) and (a – 1, 4, – 1) is parallel to the line joining the points (0, 2, b – 1) and (5, 3, – 2) find the values of a and b
Show that the points (2, 3, 4), (– 1, 4, 5) and (8, 1, 2) are collinear
If the two lines `(x - 1)/2 = (y + 1)/3 = (z - 1)/4` and `(x - 3)/1 = (y - "m")/2` = z intersect at a point, find the value of m
Show that the lines `(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 and `(x - 6)/2 = (z - 1)/3, y - 2` = 0 intersect. Aslo find the point of intersection
Show that the straight lines x + 1 = 2y = – 12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them
Find the foot of the perpendicular drawn: from the point (5, 4, 2) to the line `(x + 1)/2 = (y - 3)/3 = (z - 1)/(-1)`. Also, find the equation of the perpendicular
Choose the correct alternative:
If `vec"a", vec"b", vec"c"` are non-coplanar, non-zero vectors `[vec"a", vec"b", vec"c"]` = 3, then `{[[vec"a" xx vec"b", vec"b" xx vec"c", vec"c" xx vec"a"]]}^2` is equal to
Choose the correct alternative:
The vector equation `vec"r" = (hat"i" - hat"j" - hat"k") + "t"(6hat"i" - hat"k")` represents a straight line passing through the points
