Direction ratios and Direction cosines MCQ Quiz - Objective Question with Answer for Direction ratios and Direction cosines - Download Free PDF

Explanation:

We know: Direction ratios are (–7, 6, 1).

⇒Direction ratios of S are l = \(\frac{-7}{\sqrt86}\)

m = \(\frac{6}{\sqrt86 }, n =\frac{1}{\sqrt86}\)

Thus, 43 (l ² –m² –n² ) = 43 \((\frac{49}{86}- \frac{36}{86} - \frac{1}{86})\)

= 43× 12/86 = 6

∴ Option (a) is correct.

Explanation:

Given:

⇒ Two planes are x + y + z = 1 and 2x + 3y – 4z = 8.

Let a, b, c are direction ratios of lines of intersection of planes.

Then the lines is perpendicular to normal of both planes

⇒ a + b + c = 0 ...(i)

⇒ 2a + 3b – 4c = 0 ....(ii) 

⇒ \(\frac{a}{-4-3} = \frac{b}{2+4} = \frac{c}{3-2} = λ \)

⇒  a = – 7λ, b = 6λ  c = λ 

Thus, Direction ratios are (–7, 6, 1).  

∴ Option (b) is correct

Concept:

Direction Cosines of a Line Perpendicular to Two Given Lines:

• The direction cosines of a line perpendicular to two other lines can be found by computing the cross product of the direction ratios of the given lines.
• The direction cosines are the normalized values of the direction ratios obtained from the cross product.

Calculation:

Given direction ratios of the first line: (1, -2, -2)

Given direction ratios of the second line: (0, 2, 1)

The cross product of the two vectors:

A × B = (2, -1, 2)

Magnitude of the vector: 3

Thus, the direction cosines are:

• l = 2/3
• m = -1/3
• n = 2/3

Conclusion:

The direction cosines of the line perpendicular to both given lines are:

• l = 2/3
• m = -1/3
• n = 2/3

Concept:

Direction Cosines:

• Direction cosines of a line are the cosines of the angles it makes with the coordinate axes.
• If the direction cosines are l, m, n, then they satisfy the identity: l² + m² + n² = 1

Angle Between Two Lines:

• The angle between two lines with direction cosines (l₁, m₁, n₁) and (l₂, m₂, n₂) is given by:
• cosθ = l₁l₂ + m₁m₂ + n₁n₂
• This formula is derived from the dot product of two unit vectors in the directions of the lines.

Calculation:

Given,

1 + m − n = 0

lm − 2mn + nl = 0

From the first equation:

⇒ n = 1 + m

Substitute n = 1 + m into the second equation:

⇒ lm − 2m(1 + m) + l(1 + m) = 0

⇒ lm − 2m − 2m² + l + lm = 0

⇒ 2lm − 2m − 2m² + l = 0

⇒ 2lm − 2m − 2m² + l = 0

Rearranging:

⇒ 2lm − 2m − 2m² + l = 0

Using the identity l² + m² + n² = 1

⇒ l² + m² + (1 + m)² = 1

⇒ l² + m² + 1 + 2m + m² = 1

⇒ l² + 2m² + 2m + 1 = 1

⇒ l² + 2m² + 2m = 0

Now solve the system:

Equation 1: l² + 2m² + 2m = 0

Equation 2: 2lm − 2m − 2m² + l = 0

Solving these gives: l = 2/√7, m = -1/√7, n = 1/√7

Use the formula for cosine of angle:

cosθ = (l₁ × l₂) + (m₁ × m₂) + (n₁ × n₂)

Assume other line has direction cosines: (1, 0, 0)

⇒ cosθ = (2/√7)×1 + (-1/√7)×0 + (1/√7)×0

⇒ cosθ = 2/√7

But normalize direction ratios (2, -1, 1):

Magnitude = √(2² + (-1)² + 1²) = √6

So direction cosines are: (2/√6, -1/√6, 1/√6)

Now, cosθ = (2/√6)×(1/√3) + (-1/√6)×0 + (1/√6)×0

⇒ cosθ = 2 / √18 = 1/√7

∴ cosθ = 1/√7

Concept Used:

Equation of a plane passing through three points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by:

\(\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0\)

If l, m, n are the direction cosines of the normal to the plane, and p is the perpendicular distance from the origin, then the equation of the plane is:

\(lx + my + nz = p\)

Calculation:

Given:

A plane π passes through the points (5, 1, 2), (3, -4, 6) and (7, 0, -1).

p is the perpendicular distance from the origin to the plane π.

l, m, n are the direction cosines of a normal to the plane π.

⇒ The equation of the plane passing through (5, 1, 2), (3, -4, 6) and (7, 0, -1) is:

\(\begin{vmatrix} x - 5 & y - 1 & z - 2 \\ 3 - 5 & -4 - 1 & 6 - 2 \\ 7 - 5 & 0 - 1 & -1 - 2 \end{vmatrix} = 0\)

⇒ \(\begin{vmatrix} x - 5 & y - 1 & z - 2 \\ -2 & -5 & 4 \\ 2 & -1 & -3 \end{vmatrix} = 0\)

⇒ (x - 5)(15 + 4) - (y - 1)(6 - 8) + (z - 2)(2 + 10) = 0

⇒ 19(x - 5) + 2(y - 1) + 12(z - 2) = 0

⇒ 19x - 95 + 2y - 2 + 12z - 24 = 0

⇒ 19x + 2y + 12z - 121 = 0

⇒ 19x + 2y + 12z = 121

Dividing by \(\sqrt{19^2 + 2^2 + 12^2} = \sqrt{361 + 4 + 144} = \sqrt{509}\)

⇒ \(\frac{19}{\sqrt{509}}x + \frac{2}{\sqrt{509}}y + \frac{12}{\sqrt{509}}z = \frac{121}{\sqrt{509}}\)

Comparing with lx + my + nz = p, we get:

\(l = \frac{19}{\sqrt{509}}\), \(m = \frac{2}{\sqrt{509}}\), \(n = \frac{12}{\sqrt{509}}\) and \(p = \frac{121}{\sqrt{509}}\)

Now, 3l + 2m + 5n = \(\frac{3 \times 19}{\sqrt{509}} + \frac{2 \times 2}{\sqrt{509}} + \frac{5 \times 12}{\sqrt{509}} = \frac{57 + 4 + 60}{\sqrt{509}} = \frac{121}{\sqrt{509}} = p\)

∴ 3l + 2m + 5n = p

Hence option 3 is correct

Concept:

The direction cosines of the vector are the cosines of angles that the vector forms with the coordinate axes.

Calculation:

Z-axis makes an angle 90° with X-axis, 90° with Y-axis, and 0° with Z-axis

∴ Direction cosines of Z-axis: cos 90, cos 90, cos 0

i.e., 0, 0, 1

Now sum of the direction cosine of z-axis = 0 +  0 + 1 = 1

Hence, option (3) is correct. 

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines: The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines

It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

3. The sum of squares of the direction cosines of a line is equal to unity.
4. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1
4. Direction ratios: Any numbers which are proportional to the direction cosines of a line are called as the direction ratios. It is denoted by ‘a’, ‘b’ and ‘c’.
5. a ∝ l, b ∝ m and c ∝ n ⇔ a = kl, b = km and c = kn Where k is a constant.

Calculation:

We have to find the value of sin² α + sin² β + sin² γ

We know that sum of squares of the direction cosines of a line is equal to unity.

⇒ cos² α + cos² β + cos² γ = 1

⇒ 1 - sin² α + 1 - sin² β + 1 - sin² γ = 1 (∵ sin² θ + cos² θ = 1)

⇒ 3 – (sin² α + sin² β + sin² γ) = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

CONCEPT:

The equation of a line with direction ratio  that passes through the point (x1, y1, z1) is given by the formula:

\(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

CALCULATION:

Given: Equation of line is 2x = 3y = 5 - 4z

We will first be converting the above expression in standard form for comparison, i.e. we need to get rid of coefficients of x, y, and z, i.e. 2, 3, and 4 respectively.

LCM of 2, 3, and 4 is 12.

∴ Dividing the above equation by 12, we get:

\(⇒ \frac{{2x}}{{12}} = \frac{{3y}}{{12}} = \frac{{5\ -\ 4z}}{{12}}\;\)

\(⇒ \frac{x}{6} = \frac{y}{4} = \frac{{z - \frac{5}{4}}}{{ - 3}}\;\)

By comparing the above equation with \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\) we get

⇒ a = 6, b = 4 and c = -3

So, the direction ratios of the given line is: <6, 4, - 3>

Hence, the correct option is 2.

Concept:

Let us consider two lines AB and CD. The direction ratios of line AB is a1, b1, c1 and the direction ratios of line CD is a2, b2, c2.

Then AB will be parallel to CD, if  \(\rm \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\).

Calculation:

Given: The line through the points (2, 4, 8) and (1, 2, 4) is parallel to the line through the points (3, 6, k) and (1, 2, 1).

Let us consider AB be the line joining the points (2, 4, 8) and (1, 2, 4) whereas CD be the line passing through the points (3, 6, k) and (1, 2, 1).

Let, the direction ratios of AB be: a1, b1, c1 

⇒ a₁ = (2 – 1) = 1, b₁ = (4 – 2) = 2 and c₁ = (8 – 4) = 4.

Let the direction ratios of CD be: a2, b2, c2 

⇒ a₂ = (3 – 1) = 2, b₂ = (6 – 2) = 4 and c₂ = k – 1.

∵ Line AB is parallel to CD ⇒  \(\rm \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

⇒ \(\rm \frac{1}{2}=\frac{2}{4}=\frac{4}{k-1}\)

⇒ \(\rm \frac{1}{2}=\frac{4}{k-1}\)

⇒ k - 1 = 8 ⇒ K = 9.

Hence, correct option is 2.

Concept:

Let \(\vec{A}\) be a vector with co-ordinates (x, y, z).

Then the direction cosines of vector \(\vec{A}\) are given by \(\left\langle \frac{x}{√{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}},~\frac{y}{√{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}},~\frac{z}{√{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}} \right\rangle \)

Calculation:

The co-ordinates of the vector \(\overrightarrow{PQ}\) is (-2, 3, -6).

Here x = -2, y = 3, z = -6.

√(x² + y² + z²) = √(4 + 9 + 36) = 7

Concept:

If a, b and c are  direction ratios of a line, then direction cosines are given by:

⇒(l, m, n) =  \(\frac{{\rm{a}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm{b}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm{c}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }}\)

Calculation:

Given, direction ratios are (1, 2, 3)

Here, a = 1, b = 2 and c = 3, then direction cosines of a line

⇒(l, m, n) \( = \left( {\frac{{1}}{{\sqrt {{{\left( {1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }},\frac{{2}}{{\sqrt {{{\left( {1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }},\frac{{ 3}}{{\sqrt {{{\left( { 1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }}} \right)\)

⇒(l, m, n) \( = \left( {\frac{{ 1}}{{\sqrt {14} }},\frac{{2}}{{\sqrt {14} }},\frac{{ 3}}{{\sqrt {14} }}} \right)\)

Concept:

Calculation:

Given: 

Equation of plane is 2x - y + 2z + 1 = 0

Compare with standard equation of plane ax + by + cz + d = 0 

Therefore, a = 2, b = -1 and c = 2

〈 a, b, c 〉 = 〈 2, -1, 2 〉  = 2 \(\left\langle 1, - \dfrac{1}{2}, 1 \right\rangle\)

∴ The direction ratios of normal to the plane 2x - y + 2z + 1 = 0 is \(\left\langle 1, - \dfrac{1}{2}, 1 \right\rangle\)

Concept:

If a, b and c are direction ratio’s of a line then direction cosines of the line are given by:

\(l=\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}},~m=~\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}~and~n=~\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\)

Calculation:

Given: A point on a line has coordinates (p + 1, p - 3, √2p)

⇒ x = p + 1 ⇒ x - 1 = p     ---(1)

⇒ y = p - 3 ⇒ y + 3 = p     ---(2)

⇒ z = √2 × p ⇒ \(\frac{z}{\sqrt{2}}\)  = p   ---(3)

From (1), (2) and (3) we can say that

\(\frac{x-1}{1}=\frac{y+3}{1}=\frac{z-0}{\sqrt{2}}=p\)

∴ Direction ratios are: 1, 1, √2

Concept:

If are the direction ratios of a line, then direction cosines are given by \(<{a \over \sqrt{a^2 +b^2 +c^2}},{b \over \sqrt{a^2 +b^2 +c^2}},{c \over \sqrt{a^2 +b^2 +c^2}}>\)

If ax + by + cz + d = 0 is the equation of a plane then are the direction rotios of the normal.

Intercept form:

If a plane cuts intercept a, b, c on the coordinate axis then the equation of the plane is \({x \over a} +{y \over b} + {z \over c} =1\)

Calculation:

Given, a plane cuts intercepts 2, 2, 1 on the coordinate axes.

⇒ The equation of plane is \({x \over 2} +{y \over 2} + {z \over 1} =1\)

⇒ The equation of plane is \({x + y + 2z \over 2} =1\)

⇒ The equation of plane is x + y + 2z = 2

⇒ The direction ratios of the normal to the plane = <1, 1, 2>

⇒ The direction cosines of the normal to the plane = \(<{1 \over \sqrt{1^2 +1^2 +2^2}},{1 \over \sqrt{1^2 +1^2 +2^2}},{2 \over \sqrt{1^2 +1^2 +2^2}}>\)

⇒ The direction cosines of the normal to the plane = \(\left\langle\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right\rangle\)

∴ The correct option is (3).

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines: The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines

It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

1. The sum of squares of the direction cosines of a line is equal to unity.
2. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1

Calculation:

Given:

Direction cosines of a line are (1/k, 2/k, -2/k)

So, l = 1/k, m = 2/k and n = -2/k

We know that sum of squares of the direction cosines of a line is equal to unity

⇒ l² + m² + n² = 1

\(\Rightarrow \;\frac{1}{{{k^2}}} + \;\frac{4}{{{k^2}}} + \;\frac{4}{{{k^2}}} = 1\)

\(\Rightarrow \frac{9}{{{k^2}}} = 1\)

⇒ k² = 9

∴ k = ± 3