February 6, 2024

Explore the comprehensive NCERT Solutions for Class 12 Science Maths Chapter 5 on Three Dimensional Geometry, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these solutions prove invaluable for completing homework efficiently and preparing for exams. The Three Dimensional Geometry Solutions offer a user-friendly resource to grasp key concepts quickly. Access free answers to all questions from the NCERT Book of class 12 Science Maths Chapter 5, enhancing your understanding and aiding in academic success. Page No 467: Question 1: If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines. ANSWER: Let direction cosines of the line be l, m, and n. Therefore, the direction cosines of the line are Page No 467: Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes. ANSWER: Let the direction cosines of the line make an angle α with each of the coordinate axes. ∴ l = cos α, m = cos α, n = cos α Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are  Page No 467: Question 3: If a line has the direction ratios −18, 12, −4, then what are its direction cosines? ANSWER: If a line has direction ratios of −18, 12, and −4, then its direction cosines are Thus, the direction cosines are. Page No 467: Question 4: Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear. ANSWER: The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7). It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2), are given by, x2 − x1, y2 − y1, and z2 − z1. The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3. The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6. It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional. Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear. Page No 467: Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2) ANSWER: The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2). The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6. Therefore, the direction cosines of AB are The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4. Therefore, the direction cosines of BC are −217√, −317√, −217√-217, -317, -217The direction ratios of CA are 3−(−5), 5−(−5) and −4−(−2) i.e. 8, 10 and -2. Therefore the direction cosines of CA are 8(8)2 + (10)2 + (−2)2√, 10(8)2 + (10)2 + (−2)2√, −2(8)2 + (10)2 + (−2)2√8242√, 10242√, −2242√442√, 542√, −142√882 + 102 + -22, 1082 + 102 + -22, -282 + 102 + -228242, 10242, -2242442, 542, -142 Page No 477: Question 1: Show that the three lines with direction cosines  are mutually perpendicular. ANSWER: Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0 (i) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. (ii) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. (iii) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. Thus, all the lines are mutually perpendicular. Page No 477: Question 2: Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). ANSWER: Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). The direction ratios, a1, b1, c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4. The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4. AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0 a1a2 + b1b2+ c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4 = 6 + 10 − 16 = 0 Therefore, AB and CD are perpendicular to each other. Page No 477: Question 3: Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5). ANSWER: Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5). The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4. The direction ratios, a2, b2, c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4. AB will be parallel to CD, if  Thus, AB is parallel to CD. Page No 477: Question 4: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector. ANSWER: It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is  It is known that the line which passes through point A and parallel to is given by is a constant. This is the required equation of the line. Page No 477: Question 5: Find the equation of the line in vector and in Cartesian form that passes through the point with position vector  and is in the direction . ANSWER: It is given that the line passes through the point with position vector It is known that a line through a point with position vector and parallel to is given by the equation,  This is the required equation of the line in vector form. Eliminating λ, we obtain the Cartesian form equation as This is the required equation of the given line in Cartesian form. Page No …

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