# Important Questions for Class 12 Chapter 11 - Three dimensional Geometry

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**Important questions based on NCERT syllabus for Class 12 Chapter 11 - Three dimensional Geometry:**

*Question 1*: Find the direction of the cosines of the line (x+3)/4 = (3y + 8)/6 = (7-z)/4. Also find the vector equation of the line through the point A(-2,4,5) and parallel to the given line.

*Solution*: Given equation of the line is

(x+3)/4 = (3y + 8)/6 = (7-z)/4

This equation can be rewritten as (x+3)/4 = (y + 8/3)/2 = (z-7)/-4

So, direction ratios of the line are (4,2,-4).

Now direction cosine's of the line are

l = 4/√4^2 +2^2 +(-4)^2 , m = 2/√4^2 +2^2 +(-4)^2 and n = -4/√4^2 +2^2 +(-4)^2

Since l = +-a/√(a^2 + b^2 + c^2) , m = +-b/√(a^2 + b^2 + c^2) , n = +-c/√(a^2 + b^2 + c^2)

l = 4/√16 + 4 +16 , m = 2/√16 + 4 +16 and n = -4/√16 + 4 +16

l = 4/√36 , m = 2/√36 and n = -4/√36

l = 4/6 , m = 2/6 and n = -4/6

Hence direction cosine's of the line are 4/6, 2/6 and -4/6

Now direction of the line parallel to the given line are 4, 2, -1 and it passes through the points A(-2,4,5).

So required equation of the line parallel to the given line is

x+2/4 = y-4/2 = z-5/-4

*Question 2*: Find the angle between the lines r = i - j - k + λ(2i + 3j+ 5k) and r = (6i + 7j - 2k) + µ(7i + j + 2k)

*Solution*: a_1 = i - j + k a_2 = 6i + 7j - 6k

b_1 = 2i + 3j + 5k b_2 = 7i + j + 2k

we know that angle between two lines is given by cos theta = |b_1* b_2|/|b_1||b_2|

Therefore, cos theta = |[(6i+7j-2k)(7i + j + 2k)]/(√6^2 + 7^2 - 2^2) (√7^2 + 1^2 + 2^2)|

cos theta = (42 + 7 - 4)/(√36 + 49 - 4)(√49 + 4 + 1)

cos theta = |45/(√81√54| = |45/9√54|

cos theta = 5/√54

Hence the angle between given two lines is

theta = cos^{-1}5/√54

*Question 3*: Find the cartesian equation of the line which passes through the points (2,4, 6) and is parallel to the line x+4/4 = 6-y/6 = z - 5/9

*Solution*: Given, the required line is parallel to the line

x+4/4 = 6-y/6 = (z - 5)/9 or x+4/4 = (y - 6)/-6 = (z - 5)/9

Therefore, the direction of both lines are proportional to each other.

Hence the required equation of the line passing through (2,4,6) having direction (4,-6,9) is

x+4/4 = (y - 6)/-6 = (z - 5)/9