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NCERT Solutions Class 12 Maths Chapter-11 (Three Dimensional Geometry)Exercise 11.2

NCERT Solutions Class 12 Maths Chapter-11 (Three Dimensional Geometry)Exercise 11.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-11 (Three Dimensional Geometry)Exercise 11.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 Maths Chapter-11 (Three Dimensional Geometry)Exercise 11.2

Exercise 11.2

Q1. Show that the three lines with direction cosines 1213,313,413;413,1213,313;313,413,1213 are mutually perpendicular.

Answer.  Two lines with direction cosines, l1,m1,n1 and l2,m2,n2, are perpendicular to each  other, if I1I2+m1m2+n1n2=0 (i) For the lines with direction cosines, 1213,313,413 and 413,1213,313, we obtain l1l2+m1m2+n1n2=1213×413+(313)×1213+(413)×313=481693616912169=0 Therefore, the lines are perpendicular. (ii) For the lines with direction cosines, 413,1213,313 and 313,413,1213, we obtain l1l2+m1m2+n1n2=413×313+1213×(413)+313×1213=1216948169+36169=0 Therefore, the lines are perpendicular. Thus, all the lines are mutually perpendicular.

Q2. 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 (31),(4(1)), and (22) i.e.t 2,5, and 4 .   The direction ratios, a2,b2,c2 of co are (30),(53), and (62) i.e., 3,2, and 4. AB and CD will be perpendicular to each other, if a1a2+b1b2+c1c2=0a1a2+b1b2+c1c2=2×3+5×2+(4)×4=6+1016=0 Therefore, AB and CD are perpendicular to each other.

Q3. 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) adn CD be the through the points (-1, -2, 1) and (1, 2, 5).  The directions ratios, a1,b1,c1, of AB are (24),(37), and (48) i.e., 2,4, and 4 .   The direction ratios, a2,b2,c2, of CD are (1(1)),(2(2)), and (51) l.e., 2,4, and 4. AB will be parallel to CD, if a1a2=b1b2=c1c2 a1a2=22=1b1b2=44=1c1c2=44=1a1a2=b1b2=c1c2 Thus, AB is parallel to CD.

Q4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i^+2j^2k^.

Answer. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is a=i^+2j^+3k^ b=3i^+2j^2k^ It is known the the line which passes through point A and parallel to 0b is given by r=a+λb, where λ is a constant. r=i^+2j^+3k^+λ(3i^+2j^2k^) This is the required equation of the line.

Q5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^j+4k^ and is in the direction i^+2j^k^.

Answer.  It is given that the line passes through the point with position vector a¯=2i^+j^+4k^b=i^+2j^k^ It is known that a line through a point with position vector a and parallel to b¯ is given by   the equation, r=a+λbr=2i^j^+4k^+λ(i^+2j^k^) This is the required equation of the line in vector form. r=xi^yj^+zk^xi^yj^+zk˙=(λ+2)i^+(2λ1)j^+(λ+4)k^ 
Q6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by x+33=y45=z+86.

Answer.  It is given that the line passes through the point (2,4,5) and is parallel to x+33=y45=z+86 x+33=y45=z+86 The direction ratios of the line, y43=y45=z+86, are 3,5, and 6. The required line is parallel to x+33=y45=z+86  Therefore, its direction ratios are 3k,5k, and 6k, where k0 It is known that the equation of the line through the point (x1,y1,z1) and with direction  ratios, a,b,c, is given by xx1a=yy1b=zz1c Therefore the equation of the required line is  

Q7. The cartesian equation of a line is x53=y+47=z62. Write its vector form.

Answer.  The Cartesian equation of the line is x53=y+47=z62 The given line passes through the point (5,4,6) . The position vector of this point is a¯=5i^4j^+6k^ Also, the direction ratios of the given line are 3, 7 and 2. This means that the line is in the direction of vector b=3i^+7j^+2k^ It is known that the line through position vector b and in the direction of the vector a is given by the equation , r=a+λb,λR r=(5i^4j^+6k˙)+λ(3i^+7j^+2k^) This is the required equation of the given line in vector form.

Q8. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, -2, 3).

Answer. The required line passes through the origin. Therefore, its position vector is given by, a=0 The direction ratios of the line through origin and (5, -2, 3) are (5-0)=5,(-2-0)=-2,(3-0)=3 The line is parallel to the vector given by the equation b=5i^2j^+3k^ The equation of the line in vector through a point with position vector and parallel to b is ,r=a+λb,λR r=0+λ(5i^2j^+3k^)r=λ(5i^2j^+3k^) The equation of the line through the point (x1,y1,z1) and direction ratios a, b, c is given by xx1a=yy1b=zz1c Therefore, the equation of the required line in the Cartesian form is x05=y02=z03x5=y2=z3.
Q9. Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

Answer. Let the line passing through the points, P(3,2,5) and Q(3,2,6), be PQ. Since PQ passes through P (3,-2,-5), its position vector is given by, a=3i^2j^5k^ The direction ratios of PQ are given by, (3-3)=0,(-2+2)=0,(6+5)=11 The equation of the vector in the direction of PQ is