The image of the point 1 6 3 in the line
WebSo, the point N = (1, 3, 5) Let M (α, β, γ) be the image of P (1, 6, 3) in the given line. Then N is the midpoint of PM. α β γ ⇒ α + 1 2 = 1, β + 6 2 = 3 a n d γ + 3 2 = 5. ⇒ α = 1, β = 0 and γ = … WebReason: The unit vector perpendicular to both the lines L1 and L2 is −→ i − 7→ j + 5→ k 5√3. Q. If (a,b,c) is the image of the point (1,2,−3) in the line, x+1 2 = y−3 −2 = z −1, then a+b+c is. Q. If the distance between the plane, 23 x − 10 y − 2 z + 48 = 0 and the plane containing the lines x + 1 2 = y - 3 4 = z + 1 3 ...
The image of the point 1 6 3 in the line
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WebFree line equation calculator - find the equation of a line step-by-step WebIn this problem, we have given line. So I'm just writing the value as a vehicle to t minus one. Why is given as one plus two t and yeah, he's given us three minus 30.
Web1*3 = 3, so A' (the dilated point) should be 3 units down from P. 2*3 = 6, so A' should be 6 units to the left of P. It doesn't matter if you go left first or down first, because you always determine the location of A' with respect to P based on the location of A (which doesn't move) with respect to P. 1 comment ( 45 votes) Upvote Downvote Flag WebFeb 11, 2024 · Solution For (c) Answer any one questions : (i) Find the image of the point (1,6,3) with respect to the line 1x =2y−1 =3z−2 (ii) If a plane passes th
WebOct 30, 2024 · Let P = (1, 6, 3) and Q(λ,1 + 2λ,2 + 3λ) be two points on the line so that. Now, vector PQ is perpendicular to the given line. This implies. Therefore, the foot of the … Web1*3 = 3, so A' (the dilated point) should be 3 units down from P. 2*3 = 6, so A' should be 6 units to the left of P. It doesn't matter if you go left first or down first, because you always …
WebFeb 25, 2024 · Coordinates of image are (6,11) Explanation: As y = 4 is parallel to x -axis, abscissa of image would not change and will be 6. Further distance of (6, − 3) from y = 4 is − 3 − 4 = 7 and point is below line y = 4 hence image will be further 7 from it and hence its ordinate would be 4 +7 = 11 and Coordinates of image are (6,11)
WebStep 1.Find the image of the point We know that if B x, y is the image of a point P p, q with respect to a line a x + b y + c = 0 then x - p a = y - q b = - 2 ( ap + bq + c) a 2 + b 2 Image of the point ( 3, 5) in the line x – y + 1 = 0, is x - 3 1 = y - 5 - 1 = - 2 3 - 5 + 1 2 = 1 ∴ x = 4, y = 4 Image of the point ( 3, 5) is 4, 4 For option A baumer m310.a01WebFind the image of the point (1, 6, 3) in the line x 1 = y - 1 2 = z - 2 3. Advertisement Remove all ads Solution Let P (1, 6, 3) be the given point and let L be the foot of perpendicular … baumer m310.b06WebThe image of the point (1,6,3) in the line x 1= y−1 2 = z−2 3 is A (1,0,7) B (7,0,1) C (2,7,0) D (−1,−6,−3) Solution The correct option is A (1,0,7) Let x 1= y−1 2 = z−2 3 = k Then any … baumer lbfs datasheetWebLet the desired image point be (a, b). Then you want two conditions: 1) The midpoint of the line segment between the given point and the image point is on the given line. This gives us 2− 3 + b 2 = 3 + a 2 + 1 2) The line … dave 105.3WebSolution The correct option is C (1,0,7) Let P (1, 6, 3) be the given point and let L be the foot of the perpendicular from P to the given line. The coordinates fo a general point on the … dave 1075WebDec 25, 2015 · find the image of the point ( 1, 6, 3) in the line x 1 = y − 1 2 = z − 2 3 I want to know the general equation to find the image of a point in a line. EDIT By image, I mean a … dave 104WebThe image of my parallelogram PQRS after a reflection across WY is parallelogram P'Q'R'S'. If RR' = 12, then RZ = 6 A line segment has endpoints at (-4, -6) and (-6, 4). Which reflection will produce an image with endpoints at (4, -6) and (6, 4)? b. a reflection of the line segment across the y-axis dave 12 g3