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(1) Find the normal vectors of the lines joining the points with the following position vectors (a) 4i +7j, 6i-3j (b) -2i - 3j,
(1) Find the normal vectors of the lines joining the points with the following position vectors (a) 4i +7j, 6i-3j (b) -2i - 3j, -8 7j (c) 2i +i/2, 6+j/6 (2) Write down in normal form the equation of the line which is (a) parallel to the line r (5i+3j) = 2 and passes through the oint (2, -3); (b) perpendicular to the line 7.r+ 6y = 1 and passes through the point (1, 1); (c) passes through the points (3, 1) and (-2,-7); (3) Find the equation of the perpendicular bisector of the line join- ing the points with position vectors (a) 3i +4j, 7i +10j (b) -4i-3j, 2i - 6j. (4) In each of the following cases write down the normal vectors of the given lines and hence find the acute angles between them: (a) 2r - 5y 3 = 0, 3r +7y = 10; (b) 2r + 3y - 2 = 0, 2x- 8y+ 19 0, (c) 4y-7y+ 13 = 0, 2x- 8y+ 19 = 0 (5) Use vector methods to find the position vectors of the points of intersection of the following pairs of lines: (a) r (i-i)+2= 0, r (3i+2j) + 6 = 0; (b) r (4i - j)-3 = 0, r (i+2j) = 3; (c) 3x - y+4 =0, 3r 4y 15 = 0, (d) 7x-4y+1 = 0, r- y+1= 0. (6) Find the angles of the triangle whose sides are the lines r+2y+ 5 = 0, r-3y+4= 0 and 3r+y+7= 0. (7) Find the position vector of the centroid of the triangle formed by the lines r:(i- j) 2= 0, r (3i + j) 10 = 0, r (7i 3j) 2 = 0. (8) Find the equation of the line perpendicular to the line 4r- 3y = 0 and passes through the point of intersection of the lines r+ 2y 5 = 0, 3r - y -1 = 0.
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Qn1 a Normal vector is given by i j k 4 7 0 6 3 0 k1242 54k b Normal vector is given by i j k 2 3 0 8 7 0 k1424 10k c Normal vector is given by i j k 2 12 0 6 16 0 k26 62 83 k Qn 2 a We have r5i3j20 x...Get Instant Access to Expert-Tailored Solutions
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