Three-Dimensional Geometry PYQs Set 11 (PDF)

Set 11 of 16 · Questions 301-330 of 471

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Mere Teacher · Previous Year Questions JEE Main Mathematics · Three-Dimensional Geometry · Set 11 (Q301-330)
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301. $Let two vertices of a triangle ABC be (2, 4, 6) and (0, -2,-5). and its centroid be (2,1,-1). If the image of the third vertex in the plane x + 2y + 4z = 11 is (α, β, γ) then α β + β γ + γ α is equal to$ (1) $70$ (2) $76$ (3) $74$ (4) $72$ 302. $Consider the lines L_{1} : x - 1 = y - 2 = z and L_{2} : x - 2 = y = z - 1 . Let the feet of the perpendiculars from the point P (5, 1, - 3) on the lines L_{1} and L_{2} be Q and R respectively. If the area of the triangle PQR is A, then 4 A^{2} is equal to:$ (1) $151$ (2) $147$ (3) $139$ (4) $143$ 303. $The shortest distance between the lines \frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5} and \frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1} is:$ (1) $\frac{178}{563}$ (2) $\frac{187}{563}$ (3) $\frac{185}{563}$ (4) $\frac{179}{563}$ 304. $Equation of the perpendicular bisector of the line joining the points whose position vectors are a and b respectively is$ (1) $(2 r - a - b) \cdot (a - b) = 0$ (2) $(2 r - a - b) \cdot (a + b) = 0$ (3) $(2 r + a + b) \cdot (a - b) = 0$ (4) $(2 r - a + b) \cdot (a + b) = 0$ 305. $The distance of the point having position vector i - 2 j - 6 k, from the straight line passing through the point (2, - 3, - 4) and parallel to the vector 6 i + 3 j - 4 k is units.$ (1) $\frac{340}{61}$ (2) $\frac{341}{61}$ (3) $\frac{341}{61}$ (4) $\frac{341}{61}$ 306. $Let (α, β, γ) be mirror image of the point (2, 3, 5) in the line \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} . Then 2 α + 3 β + 4 γ is equal to$ (1) $32$ (2) $33$ (3) $31$ (4) $34$ 307. $Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x - 3y + 5z = 8. If the mirror image of the point (2, - \frac{1}{2}, 2) in the rotated plane is B(a, b, c), then :$ (1) $\frac{a}{8} = \frac{b}{5} = \frac{c}{- 4}$ (2) $\frac{a}{4} = \frac{b}{5} = \frac{c}{- 2}$ (3) $\frac{a}{8} = \frac{b}{- 5} = \frac{c}{4}$ (4) $\frac{a}{4} = \frac{b}{5} = \frac{c}{2}$ 308. $Let the plane passing through the point (-1, 0,-2) and perpendicular to each of the planes 2x + y - z = 2 and x - y - z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to:$ (1) $3$ (2) $8$ (3) $5$ (4) $4$ 309. $A perpendicular is drawn from a point on the line \frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{1} to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x - y + z = 3 Then the co-ordinates of Q are$ (1) $(1, 0, 2)$ (2) $(2, 0, 1)$ (3) $(- 1, 0, 4)$ (4) $(4, 0, - 1)$ 310. $Let (α, β, γ) be the image of point P (2, 3,5) in the plane 2x + y - 3z = 6. Then α + β + γ equal to$ (1) $5$ (2) $10$ (3) $12$ (4) $9$ 311. $Let the points on the plane P be equidistant from the points (- 4, 2, 1) and (2, - 2, 3) . Then the acute angle between the plane P and the plane 2 x + y + 3 z = 1 is$ (1) $\frac{π}{6}$ (2) $\frac{π}{4}$ (3) $\frac{π}{3}$ (4) $\frac{5 π}{12}$ 312. $The plane, passing through the points (0, -1, 2) and (-1, 2, 1) and parallel to the line passing through (5, 1, -7) and (1, -1, -1), also passes through the point$ (1) $(-2, 5, 0)$ (2) $(1, - 2, 1)$ (3) $(2, 0, 1)$ (4) $(0, 5, -2)$ 313. $A plane P meets the coordinate axes at A, B and C respectively. The centroid of Δ ABC is given to be (1, 1, 2) . Then the equation of the line through this centroid and perpendicular to the plane P is:$ (1) $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$ (2) $\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 2}{2}$ (3) $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ (4) $\frac{x - 1}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$ 314. $Let the foot of the perpendicular from the point (1, 2, 4) on the line \frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3} be P. Then the distance of P from the plane 3x+4y+12z+23=0$ (1) $5$ (2) $\frac{50}{13}$ (3) $4$ (4) $\frac{63}{13}$ 315. $If (a, b, c) is the image of the point (1, 2, - 3) in the line, \frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} then a + b + c is equal to:$ (1) $2$ (2) $1$ (3) $3$ (4) $- 1$ 316. $If the shortest distance between the lines L_{1} : r = (2 + λ) i + (1 - 3 λ) j + (3 + 4 λ) k, λ \in R L_{2} : r = 2 (1 + μ) i + 3 (1 + μ) j + (5 + μ) k, μ \in R is \frac{m}{n}, where gcd (m, n) = 1, then the value of m + n equals$ (1) $390$ (2) $384$ (3) $377$ (4) $387$ 317. $If the equation of the plane containing the line x + 2 y + 3 z - 4 = 0 = 2 x + y - z + 5 and perpendicular to the plane r = (i^- j^) + λ (i^+ j^+ k^) + μ ∣ (i^- 2 j^+ 3 k^) is a x + b y + cz = 4 then (a - b + c)$ (1) $18$ (2) $22$ (3) $20$ (4) $24$ 318. $Let A be the point of intersection of the lines L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1} and L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5} . Let B and C be the points on the lines L_{1} and L_{2} respectively such that AB = AC = 15 . Then the square of the area of the triangle ABC is:$ (1) $63$ (2) $57$ (3) $60$ (4) $54$ 319. $Let P be the foot of the perpendicular from the point Q (10, - 3, - 1) on the line \frac{x - 3}{7} = \frac{y - 2}{- 1} \frac{z + 1}{- 2} . Then the area of the right angled triangle PQR, where R is the point (3, - 2, 1), is$ (1) $915$ (2) $30$ (3) $815$ (4) $330$ 320. $The foot of the perpendicular drawn from the origin to the plane is (4, - 2, 5), then the cartesian equation of the plane is$ (1) $4 x - 2 y + 5 z = 45$ (2) $- 4 x + 2 y + 5 z = 45$ (3) $4 x - 2 y + 5 z + 45 = 0$ (4) $4 x + 2 y - 5 z + 45 = 0$ 321. $Let the line passing through the points (- 1, 2, 1) and parallel to the line \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} intersect the line \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} at the point P . Then the distance of P from the point Q (4, - 5, 1) is$ (1) $5$ (2) $55$ (3) $56$ (4) $10$ 322. $Let the vertices Q and R of the triangle PQR lie on the line \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}, QR = 5 and the coordinates of the point P be (0, 2, 3) . If the area of the triangle PQR is \frac{m}{n} then:$ (1) $2 m - 521 n = 0$ (2) $m - 521 n = 0$ (3) $5 m - 212 n = 0$ (4) $5 m - 221 n = 0$ 323. $Let a line passing through the point (4, 1, 0) intersect the line L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} at the point A (α, β, γ) and the line L_{2} : x - 6 = y = - z + 4 at the point B (a, b, c) . Then 1 α a 0 β b 1 γ c is equal to$ (1) $16$ (2) $6$ (3) $8$ (4) $12$ 324. $Let P be the point of intersection of the lines \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} and \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} . Then, the shortest distance of P from the line 4 x = 2 y = z is$ (1) $\frac{5 14}{7}$ (2) $\frac{3 14}{7}$ (3) $\frac{14}{7}$ (4) $\frac{6 14}{7}$ 325. $If the mirror image of the point P (3, 4, 9) in the line \frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1} is (α, β, γ), then 14 (α + β + γ) is :$ (1) $102$ (2) $138$ (3) $108$ (4) $132$ 326. $Let (α, β, γ) be the image of the point (8, 5, 7) in the line \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5} . Then α + β + γ is equal to :$ (1) $16$ (2) $20$ (3) $14$ (4) $18$ 327. $Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)^{2} is equal to .$ 328. $Let P be the image of the point Q (7, - 2, 5) in the line L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} and R (5, p, q) be a point on L . Then the square of the area of △ PQR is .$ 329. $Let P_{1} : r . (2 i^+ j^- 3 k^) = 4 be a plane. Let P_{2} be another plane which passes through the points (2, -3, 2) (2, - 2, - 3) and (1, -4, 2). If the direction ratios of the line of intersection of P_{1} and P_{2} be 16, α, β, then the value of α + β is equal to_____.$ 330. $The plane passing through the line L: l x-y+3(l- l) z = 1, x+2y - z = 2 and perpendicular to the plane 3x+2y+z = 6 is 3x-8y+7z=4. If θ is the acute angle between the line L and the y-axis, then 415 cos^{2} θ is equal to_____.$