Question 1

The acute angle between the planes 2 x - y + z = 6 and x + y + 2 z = 7 is

Accepted Answer

60°

Question 2

A line makes angle α, β, γ with x-axis, y-axis and z-axis respectively then cos 2α + cos 2β + cos 2γ is equal to

Accepted Answer

-1

Question 3

The distance of point (2, 5, 7) from the x-axis is

Accepted Answer

√74

Question 4

If a line makes angles $90^{\circ}$, $135^{\circ}$, $45^{\circ}$ with the $x, y$ and $z$-axes respectively, find its direction cosines.

Accepted Answer

The direction cosines of a line are the cosines of the angles it makes with the coordinate axes. Given the angles are 90°, 135°, and 45° with x, y, and z axes respectively, the direction cosines are:

l = cos 90° = 0
m = cos 135° = -\frac{1}{\sqrt{2}}
 n = cos 45° = \frac{1}{\sqrt{2}}

Hence, the direction cosines are (0, -1/\sqrt{2}, 1/\sqrt{2}). — Direction cosines are defined as the cosines of the angles made by the line with the x, y, and z axes. Substitute the given angles to find each direction cosine.

Question 5

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Accepted Answer

Let the equal angle be 	heta. Since the line makes equal angles with the x, y, and z axes, the direction cosines are all equal:

l = m = n = \cos 	heta

Since the sum of the squares of direction cosines is 1:

l^2 + m^2 + n^2 = 1
3l^2 = 1
l = \pm \frac{1}{\sqrt{3}}

Therefore, the direction cosines are \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}ight) or \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}ight). — Since the line makes equal angles with all axes, direction cosines are equal. Using the property that the sum of squares of direction cosines equals 1, solve for each direction cosine.

Question 6

If a line has the direction ratios $-18, 12, -4$, then what are its direction cosines?

Accepted Answer

Given direction ratios are a = -18, b = 12, c = -4.

The direction cosines are given by:

l = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = \frac{-18}{\sqrt{(-18)^2 + 12^2 + (-4)^2}} = \frac{-18}{\sqrt{324 + 144 + 16}} = \frac{-18}{\sqrt{484}} = \frac{-18}{22} = -\frac{9}{11}

m = \frac{12}{22} = \frac{6}{11}

n = \frac{-4}{22} = -\frac{2}{11}

Therefore, the direction cosines are \left(-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}\right). — Direction cosines are the normalized direction ratios. Calculate the magnitude of the vector formed by the direction ratios and divide each ratio by this magnitude.

Question 7

Show that the points $(2, 3, 4), (-1, -2, 1), (5, 8, 7)$ are collinear.

Accepted Answer

To show that points A(2,3,4), B(-1,-2,1), and C(5,8,7) are collinear, check if the vectors AB and BC are scalar multiples.

Vector AB = B - A = (-1 - 2, -2 - 3, 1 - 4) = (-3, -5, -3)
Vector BC = C - B = (5 - (-1), 8 - (-2), 7 - 1) = (6, 10, 6)

Check if AB = k * BC for some scalar k:

Comparing components:
-3 = 6k => k = -\frac{1}{2}
-5 = 10k => k = -\frac{1}{2}
-3 = 6k => k = -\frac{1}{2}

Since k is the same for all components, vectors AB and BC are collinear.

Therefore, points A, B, and C are collinear. — If vectors AB and BC are scalar multiples, then points A, B, and C lie on the same straight line, proving collinearity.

Question 8

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2)$.

Accepted Answer

Let the vertices be A(3,5,-4), B(-1,1,2), and C(-5,-5,-2).

Find direction cosines of sides AB, BC, and CA.

Vector AB = B - A = (-1 - 3, 1 - 5, 2 - (-4)) = (-4, -4, 6)
Magnitude of AB = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}

Direction cosines of AB:

l = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}

m = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}

n = \frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}

Vector BC = C - B = (-5 - (-1), -5 - 1, -2 - 2) = (-4, -6, -4)
Magnitude of BC = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}

Direction cosines of BC:

l = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}

m = \frac{-6}{2\sqrt{17}} = -\frac{3}{\sqrt{17}}

n = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}

Vector CA = A - C = (3 - (-5), 5 - (-5), -4 - (-2)) = (8, 10, -2)
Magnitude of CA = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}

Direction cosines of CA:

l = \frac{8}{2\sqrt{42}} = \frac{4}{\sqrt{42}}

m = \frac{10}{2\sqrt{42}} = \frac{5}{\sqrt{42}}

n = \frac{-2}{2\sqrt{42}} = -\frac{1}{\sqrt{42}}

Hence, the direction cosines of sides AB, BC, and CA are as above. — Calculate the vectors representing the sides by subtracting coordinates of vertices, then find their magnitudes and divide each component by the magnitude to get direction cosines.

Question 9

Show that the three lines with direction cosines

\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} are mutually perpendicular.

Accepted Answer

To show that the three lines are mutually perpendicular, we need to verify that the dot product of the direction cosines of each pair of lines is zero.

Let the direction cosines of the three lines be:

Line 1: (12/13, -3/13, -4/13)
Line 2: (4/13, 12/13, 3/13)
Line 3: (3/13, -4/13, 12/13)

Check dot product of Line 1 and Line 2:
= (12/13)*(4/13) + (-3/13)*(12/13) + (-4/13)*(3/13)
= (48/169) + (-36/169) + (-12/169) = 0

Check dot product of Line 2 and Line 3:
= (4/13)*(3/13) + (12/13)*(-4/13) + (3/13)*(12/13)
= (12/169) + (-48/169) + (36/169) = 0

Check dot product of Line 3 and Line 1:
= (3/13)*(12/13) + (-4/13)*(-3/13) + (12/13)*(-4/13)
= (36/169) + (12/169) + (-48/169) = 0

Since all dot products are zero, the three lines are mutually perpendicular. — The direction cosines of two lines are perpendicular if their dot product is zero. Calculating the dot products pairwise and obtaining zero confirms mutual perpendicularity.

Question 10

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  $(3, 5, 6)$ .

Accepted Answer

Find direction vectors of both lines.

Line 1 direction vector = (3 - 1, 4 - (-1), -2 - 2) = (2, 5, -4)
Line 2 direction vector = (3 - 0, 5 - 3, 6 - 2) = (3, 2, 4)

Dot product = 2*3 + 5*2 + (-4)*4 = 6 + 10 - 16 = 0

Since dot product is zero, the lines are perpendicular. — Two lines are perpendicular if their direction vectors are perpendicular, i.e., their dot product is zero.

Question 11

Show that the line through the points  $(4, 7, 8)$ ,  $(2, 3, 4)$  is parallel to the line through the points  $(-1, -2, 1)$ ,  $(1, 2, 5)$ .

Accepted Answer

Find direction vectors of both lines.

Line 1 direction vector = (2 - 4, 3 - 7, 4 - 8) = (-2, -4, -4)
Line 2 direction vector = (1 - (-1), 2 - (-2), 5 - 1) = (2, 4, 4)

Check if one vector is scalar multiple of the other:
(-2, -4, -4) = -1 * (2, 4, 4)

Since direction vectors are scalar multiples, lines are parallel. — Lines are parallel if their direction vectors are scalar multiples of each other.

Question 12

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

Accepted Answer

The vector form of the line passing through point $ (1,2,3) $ and parallel to vector $ 3\hat{i} + 2\hat{j} - 2\hat{k} $ is:

$$ \vec{r} = \vec{a} + \lambda \vec{b} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 2\hat{k}) $$

Cartesian form:

$$ \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2} $$ — Equation of line passing through point $ \vec{a} $ and parallel to vector $ \vec{b} $ is $ \vec{r} = \vec{a} + \lambda \vec{b} $.

Question 13

Find the equation of the line in vector and in cartesian form that passes through the point with position vector  $2\hat{i} - j + 4\hat{k}$  and is in the direction  $\hat{i} + 2\hat{j} - \hat{k}$ .

Accepted Answer

Given point: $ \vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} $

Direction vector: $ \vec{b} = \hat{i} + 2\hat{j} - \hat{k} $

Vector form:

$$ \vec{r} = \vec{a} + \lambda \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k}) $$

Cartesian form:

$$ \frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1} $$ — Equation of line passing through point $ \vec{a} $ and parallel to vector $ \vec{b} $ is $ \vec{r} = \vec{a} + \lambda \vec{b} $.

Question 14

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

Accepted Answer

Direction ratios of given line are (3, 5, 6).

Line passing through $(-2,4,-5)$ and parallel to given line has same direction ratios.

Cartesian equation:

$$ \frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6} $$ — Lines parallel have same direction ratios. Use point and direction ratios to write equation.

Question 15

The cartesian equation of a line is  $\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}$ . Write its vector form.

Accepted Answer

Given line:

$$ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} $$

Vector form:

$$ \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}) $$ — Vector form is $ \vec{r} = \vec{a} + \lambda \vec{b} $, where $ \vec{a} $ is position vector of point on line and $ \vec{b} $ is direction vector.

Three Dimensional Geometry

Three Dimensional Geometry — Study Notes

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

11.2.1 Direction cosines of a line passing through two points

Practice Questions — Three Dimensional Geometry

All 7 Chapters in Mathematics Part-II