Three-Dimensional Geometry PYQs Set 9 (PDF)

Set 9 of 16 · Questions 241-270 of 471

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Mere Teacher · Previous Year Questions JEE Main Mathematics · Three-Dimensional Geometry · Set 9 (Q241-270)
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241. $The direction cosines of the vector a = - 2 i + j - 5 k are$ (1) $\frac{- 2}{8}, \frac{1}{8}, \frac{- 5}{8}$ (2) $\frac{- 2}{30}, \frac{1}{30}, \frac{- 5}{30}$ (3) $\frac{2}{8}, \frac{- 1}{8}, \frac{5}{8}$ (4) $\frac{- 2}{30}, \frac{- 1}{30}, \frac{- 5}{30}$ 242. $The locus of a point P such that PA + PB = 4 where A (2, 3, 4), B (- 2, 3, 4) is$ (1) $y^{2} + z^{2} + 6 y + 8 z + 25 = 0$ (2) $y^{2} - z^{2} + 6 y + 8 z - 25 = 0$ (3) $y^{2} + z^{2} - 6 y - 8 z + 25 = 0$ (4) $y^{2} + z^{2} - 6 y - 8 z - 25 = 0$ 243. $If the plane 2x + y - 5z = 0 is rotated about its line of intersection with the plane 3x-y + 4z-7 = 0 by an angle of \frac{π}{2}, then the plane after the rotation passes through the point:$ (1) $(2, -2, 0)$ (2) $(-2, 2, 0)$ (3) $(l,0,2)$ (4) $(-l,0,-2)$ 244. $A plane passing through the point (3,1,1) contains two lines whose direction ratios are 1,-2,2 and 2,3,-1 respectively. If this plane also passes through the point (α, - 3, 5), then α is equal to:$ (1) $5$ (2) $10$ (3) $- 10$ (4) $- 5$ 245. $If the direction cosines of two lines are (\frac{2}{3}, \frac{2}{3}, \frac{1}{3}) and (\frac{5}{13}, \frac{12}{13}, 0), then identify the direction ratios of a line which is bisecting one of the angle between them.$ (1) $(40, 60, 13)$ (2) $(41, 60, 10)$ (3) $(41, 62, 13)$ (4) $(1, 2, 3)$ 246. $The point of intersection of the line passing through the points i - j, j - k and the plane passing through the points 2 i + j, 2 j - k, i + 2 k is$ (1) $\frac{1}{6} (- 5 i + 16 j - 11 k)$ (2) $\frac{1}{23} (22 i - 44 j + 25 k)$ (3) $\frac{1}{5} (18 i + 16 j - 21 k)$ (4) $\frac{1}{11} (5 i - 41 j + 21 k)$ 247. $Reflection of the point (α, β, γ) in XY plane is$ (1) $(0, 0, γ)$ (2) $(α, β, - γ)$ (3) $(- α, - β, γ)$ (4) $(α, β, 0)$ 248. $Consider the line L given by the equation \frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1} . Let Q be the mirror image of the point (2, 3,-1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?$ (1) $(-1, 1, 2)$ (2) $(1, 1, 1)$ (3) $(1, 1, 2)$ (4) $(1, 2, 2)$ 249. $Equation of the plane perpendicular to the line \frac{x}{1} = \frac{y}{2} = \frac{z}{3} and passing through the point (2, 3, 4) is$ (1) $2 x + 3 y + z = 17$ (2) $x + 2 y + 3 z = 9$ (3) $3 x + 2 y + z = 16$ (4) $x + 2 y + 3 z = 20$ 250. $Let the shortest distance between the lines L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0 and L_{1} : x + 1 = y - 1 ‘ = 4 - z be 26 . If (α, β, γ) lies on L, then which of the following is NOT possible.$ (1) $α + 2 γ = 24$ (2) $2 α + γ = 7$ (3) $2 α - γ = 9$ (4) $α - 2 γ = 19$ 251. $A variable plane is at a distance of 6 units from the origin. If it meets the coordinate axes in A, B and C, then the equation of the locus of the centroid of the △ ABC is$ (1) $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = \frac{1}{4}$ (2) $x^{2} + y^{2} + z^{2} = 4$ (3) $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 1$ (4) $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} - \frac{1}{z ^{2}} = \frac{1}{4}$ 252. $If the cartesian equation of a line is 6 x - 2 = 3 y + 1 = 2 z - 2, then the vector equation of the line is$ (1) $r = (\frac{1}{3} i - \frac{1}{3} j + k) + λ (i + 2 j + 3 k)$ (2) $r = (i + j + k) + λ (i + 2 i + 3 k)$ (3) $r = (\frac{- 1}{3} i + \frac{1}{3} j + k) + λ (i - 2 j + 3 k)$ (4) $r = (\frac{1}{3} i - \frac{1}{3} j - k) + λ (i - j + k)$ 253. $If the square of the shortest distance between the lines \frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3} and \frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5} is \frac{m}{n}, where m, n are coprime numbers, then m + n is equal to:$ (1) $21$ (2) $9$ (3) $14$ (4) $6$ 254. $If the plane 56 x + 4 y + 9 z = 2016 meets the coordinate axes in A, B and C , then the centroid of the △ ABC is$ (1) $(12, 168, 224)$ (2) $(12, 168, 112)$ (3) $(12, 168, \frac{224}{3})$ (4) $(12, - 168, \frac{224}{3})$ 255. $The shortest distance between the lines x = y = z and x = 1 - y = \frac{z}{0} is$ (1) $\frac{1}{2}$ (2) $\frac{1}{2}$ (3) $\frac{1}{3}$ (4) $\frac{1}{6}$ 256. $Let a, b and c be any three non-coplanar vectors. If m , n are scalars such that a + b = m d - c and b + c = na - d, then 3 a + 2 b + 2 c + d =$ (1) $a - d$ (2) $a + d$ (3) $0$ (4) $b + c + 2 d$ 257. $Let A (1, 2, 3), B (- 1, 4, 6), C (0, - 6, 4) and D (1, 1, 1) be the vertices of a tetrahedron, G be its centroid and G_{1} be the centroid of its face BCD . Then \frac{A G _{1}}{AG} =$ (1) $\frac{5}{3}$ (2) $\frac{4}{3}$ (3) $\frac{7}{6}$ (4) $\frac{5}{4}$ 258. $On which of the following lines lies the point of inter-section of the line, \frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1} and the plane, x + y + z = 2$ (1) $\frac{x + 3}{3} = \frac{4 - y}{3} = \frac{z + 1}{- 2}$ (2) $\frac{x - 4}{1} = \frac{y - 5}{1} = \frac{z - 5}{- 1}$ (3) $\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z + 4}{- 5}$ (4) $\frac{x - 2}{2} = \frac{y - 3}{2} = \frac{z + 3}{3}$ 259. $Let the values of p, for which the shortest distance between the lines \frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5} and r = (p i + 2 j + k) + λ (2 i + 3 j + 4 k) is \frac{1}{6}, be a, b, (a < b) . Then the length of the latus rectum of the ellipse \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 is:$ (1) $\frac{3}{2}$ (2) $9$ (3) $18$ (4) $\frac{2}{3}$ 260. $If the length of the perpendicular from the point (β, 0, β) (β \neq = 0) to the line, \frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1} is \frac{3}{2} then β is equal to:$ (1) $1$ (2) $2$ (3) $- 1$ (4) $- 2$ 261. $Consider the lines L_{1} and L_{2} given by L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} L_{2} : \frac{x - 2}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} A line L_{1} having direction ratios 1, - 1, - 2 intersects L_{1} and L_{2} at the points P and Q respectively. Then the length of line segment PQ is$ (1) $26$ (2) $32$ (3) $43$ (4) $4$ 262. $The point on the plane 2 x - 2 y + 4 z + 5 = 0 that is nearer to (1, \frac{3}{2}, 2) is$ (1) $(0, \frac{5}{2}, 0)$ (2) $(- 5, \frac{- 5}{2}, 0)$ (3) $(0, 0, \frac{- 5}{4})$ (4) $(- \frac{1}{2}, 0, - 1)$ 263. $The sum of all values of α, for which the shortest distance between the lines \frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α} is 2, is$ (1) $6$ (2) $8$ (3) $-8$ (4) $-6$ 264. $Let u and v be two vectors in a plane. Then any vector w in the plane can be written as w - a u + b v for some scalars ' a' and ' b' if and only if$ (1) $None of u and v is a scalar multiple of the other$ (2) $None of \| u \| and \| v \| is a scalar multiple of the other$ (3) $u and v have different direction$ (4) $u and v are perpendicular to each other$ 265. $Let (α, β, γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line r = (- i + 3 j + k) + λ (2 i + 3 j - k) . Then the length of the projection of the vector α i + β j + γ k on the vector 6 i + 2 j + 3 k is :$ (1) $\frac{15}{7}$ (2) $4$ (3) $\frac{18}{7}$ (4) $3$ 266. $Acute angle between the line \frac{x - 5}{2} = \frac{y + 1}{- 1} = \frac{z + 4}{1} and the plane 3 x - 4 y - z + 5 = 0 is$ (1) $cos^{- 1} (\frac{9}{364})$ (2) $cos^{- 1} (\frac{5}{2 13})$ (3) $sin^{- 1} (\frac{9}{364})$ (4) $sin^{- 1} (\frac{5}{2 13})$ 267. $Let π_{1} be the plane passing through the point 2 i - j + k and perpendicular to the vector ai + 2 j - 3 k and π_{2} be the plane passing through the point i + 2 j - k and perpendicular to the vector i - 2 j + k . If θ is the angle between the planes π_{1} and π_{2} and cos θ = - \frac{3}{7}, then the integral value of a is$ (1) $-2$ (2) $-1$ (3) $2$ (4) $1$ 268. $The length of the perpendicular drawn from the point (1, 2, 3) to the line \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2} is$ (1) $4 units$ (2) $5 units$ (3) $6 units$ (4) $7 units$ 269. $If λ_{1} < λ_{2} are two values of λ Such that the angle between the planes P_{1} : r (3 i - 5 j + k) = 7 and P_{2} : r (λi + j - 3 k) = 9 is sin^{- 1} (\frac{2 6}{5}), then the square of the length of perpendicular from the point (38 λ_{1}, 10 λ_{2}, 2) to the plane P_{1} is$ 270. $The square of the distance of the point of intersection of the line \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6} and the plane 2x - y + z = 6 from the point (-1,-1, 2) is .$