Exercise 11.2 — Three Dimensional Geometry

1. 13 marksNCERT Exercise

   Show that the three lines with direction cosines 12/13, -3/13, -4/13; 4/13, 12/13, 3/13; 3/13, -4/13, 12/13 are mutually perpendicular.

   Hint. Two lines are perpendicular exactly when the dot product of their direction cosines is zero — check all three pairs.

   Since the pairwise dot products are (12)(4)+(-3)(12)+(-4)(3)=48-36-12=0, (4)(3)+(12)(-4)+(3)(12)=12-48+36=0, and (12)(3)+(-3)(-4)+(-4)(12)=36+12-48=0 (each divided by 169), all three pairs are perpendicular.

   ✦ Mutually perpendicular, since all three pairwise dot products are zero

2. 23 marksNCERT Exercise

   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).

   Hint. Find each line's direction ratios by subtracting endpoints, then check their dot product is zero.

   Since the first line has direction ratios (2,5,-4) and the second has (3,2,4), their dot product is 6+10-16=0, which means the two lines are perpendicular.

   ✦ Perpendicular, since (2)(3)+(5)(2)+(-4)(4)=0

3. 33 marksNCERT Exercise

   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).

   Hint. Find each line's direction ratios and check whether one set is a scalar multiple of the other.

   Since the first line has direction ratios (-2,-4,-4) and the second has (2,4,4), and the first is exactly -1 times the second, the two lines are parallel.

   ✦ Parallel, since the direction ratios are proportional (in the ratio -1:1)

4. 43 marksNCERT Exercise

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

   Hint. Use the vector form r = a + lambda.b, where a is the position vector of the point and b the direction vector.

   Since the line passes through the point with position vector i+2j+3k and is parallel to 3i+2j-2k, substituting directly into r = a + lambda.b gives the vector equation.

   ✦ r = (i+2j+3k) + lambda(3i+2j-2k)

5. 53 marksNCERT Exercise

   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.

   Hint. Use r = a + lambda.b for the vector form, then extract the cartesian form from the point and direction ratios.

   Since a=(2,-1,4) and b=(1,2,-1), the vector form is r=(2i-j+4k)+lambda(i+2j-k); because the cartesian form uses the point coordinates and direction ratios directly, it becomes (x-2)/1=(y+1)/2=(z-4)/(-1).

   ✦ Vector: r=(2i-j+4k)+lambda(i+2j-k). Cartesian: (x-2)/1=(y+1)/2=(z-4)/(-1)

6. 63 marksNCERT Exercise

   Find the cartesian equation of the line which passes through the point (-2,4,-5) and is parallel to the line given by (x+3)/3=(4-y)/5=(z+8)/6.

   Hint. Extract the direction ratios from the given line first, watching the sign on the y-term.

   Since (4-y)/5 can be rewritten as (y-4)/(-5), the given line's direction ratios are (3,-5,6); because parallel lines share direction ratios, the required line through (-2,4,-5) uses the same ones.

   ✦ (x+2)/3 = (y-4)/(-5) = (z+5)/6

7. 73 marksNCERT Exercise

   The cartesian equation of a line is (x-5)/3=(y+4)/7=(z-6)/2. Write its vector form.

   Hint. Read the point and direction ratios directly off the cartesian equation.

   Since the cartesian form gives the point (5,-4,6) and direction ratios (3,7,2), substituting these into r = a + lambda.b produces the vector equation directly.

   ✦ r = (5i-4j+6k) + lambda(3i+7j+2k)

8. 83 marksNCERT Exercise

   Find the angle between the following pairs of lines: (i) r=2i-5j+k+lambda(3i+2j+6k) and r=7i-6k+mu(i+2j+2k) (ii) r=3i+j-2k+lambda(i-j-2k) and r=2i-j-56k+mu(3i-5j-4k).

   Hint. Use cos(theta)=|b1.b2|/(|b1||b2|) with the two direction vectors.

   For (i), the direction vectors are (3,2,6) and (1,2,2), giving |b1.b2|=|3+4+12|=19 and magnitudes 7 and 3, so cos(theta)=19/21. For (ii), the direction vectors are (1,-1,-2) and (3,-5,-4), giving |3+5+8|=16 and magnitudes sqrt6 and sqrt50, so cos(theta)=16/sqrt300.

   ✦ (i) cos^-1(19/21) (ii) cos^-1(16/sqrt(300)) = cos^-1(8/(5.sqrt(3)))

9. 93 marksNCERT Exercise

   Find the angle between the following pair of lines: (i) (x-2)/2=(y-1)/5=(z+3)/(-3) and (x+2)/(-1)=(y-4)/8=(z-5)/4 (ii) x/2=y/2=z/1 and (x-5)/4=(y-2)/1=(z-3)/8.

   Hint. Use cos(theta)=|a1a2+b1b2+c1c2|/(sqrt(a1^2+b1^2+c1^2).sqrt(a2^2+b2^2+c2^2)).

   For (i), the direction ratios (2,5,-3) and (-1,8,4) give |−2+40−12|=26 and magnitudes sqrt38 and sqrt81=9, so cos(theta)=26/(9.sqrt38). For (ii), the direction ratios (2,2,1) and (4,1,8) give |8+2+8|=18 and magnitudes 3 and 9, so cos(theta)=18/27=2/3.

   ✦ (i) cos^-1(26/(9.sqrt(38))) (ii) cos^-1(2/3)

10. 103 marksNCERT Exercise

    Find the values of p so that the lines (1-x)/3=(7y-14)/(2p)=(z-3)/2 and (7-7x)/(3p)=(y-5)/1=(6-z)/5 are at right angles.

    Hint. Rewrite each equation in standard form to extract clean direction ratios, then set their dot product to zero.

    Rewriting the first line gives direction ratios (-3, 2p/7, 2) and the second gives (-3p/7, 1, -5); since perpendicularity requires the dot product to vanish, solving 9p/7 + 2p/7 - 10 = 0 gives the value of p.

    ✦ p = 70/11

11. 113 marksNCERT Exercise

    Show that the lines (x-5)/7=(y+2)/(-5)=z/1 and x/1=y/2=z/3 are perpendicular to each other.

    Hint. Extract both sets of direction ratios and check their dot product.

    Since the direction ratios are (7,-5,1) and (1,2,3), their dot product is 7-10+3=0, which confirms the two lines are perpendicular.

    ✦ Perpendicular, since (7)(1)+(-5)(2)+(1)(3)=0

12. 123 marksNCERT Exercise

    Find the shortest distance between the lines r=(i+2j+k)+lambda(i-j+k) and r=(2i-j-k)+mu(2i+j+2k).

    Hint. Use the skew-lines formula: SD = |(b1 x b2).(a2-a1)| / |b1 x b2|.

    Since b1 x b2 = (-3,0,3) with magnitude 3sqrt2, and a2-a1 = (1,-3,-2), the numerator |(-3)(1)+(0)(-3)+(3)(-2)| = 9, so the shortest distance is 9/(3sqrt2).

    ✦ 3/sqrt(2) units

13. 133 marksNCERT Exercise

    Find the shortest distance between the lines (x+1)/7=(y+1)/(-6)=(z+1)/1 and (x-3)/1=(y-5)/(-2)=(z-7)/1.

    Hint. Convert both to vector form, then apply the skew-lines shortest-distance formula.

    With a1=(-1,-1,-1), b1=(7,-6,1), a2=(3,5,7), b2=(1,-2,1), computing b1 x b2 and taking |(b1 x b2).(a2-a1)|/|b1 x b2| gives the shortest distance.

    ✦ 2.sqrt(29) units

14. 143 marksNCERT Exercise

    Find the shortest distance between the lines whose vector equations are r=(i+2j+3k)+lambda(i-3j+2k) and r=(4i+5j+6k)+mu(2i+3j+k).

    Hint. Apply the skew-lines formula directly with the given points and direction vectors.

    With a1=(1,2,3), b1=(1,-3,2), a2=(4,5,6), b2=(2,3,1), computing the cross product b1 x b2 and applying |(b1 x b2).(a2-a1)|/|b1 x b2| gives the result.

    ✦ 3/sqrt(19) units

15. 153 marksNCERT Exercise

    Find the shortest distance between the lines whose vector equations are r=(1-t)i+(t-2)j+(3-2t)k and r=(s+1)i+(2s-1)j-(2s+1)k.

    Hint. Rewrite each in the standard r=a+lambda.b form first — check whether the direction vectors turn out parallel.

    Rewriting gives a1=(1,-2,3), b1=(-1,1,-2) and a2=(1,-1,-1), b2=(1,2,-2); since these direction vectors are not proportional, the lines are skew, so applying the standard skew formula gives the shortest distance.

    ✦ 8/sqrt(29) units

Solutions written by the tuition.in editorial team and checked against lemh205.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.