Perpendicular Lines MCQ Quiz - Objective Question with Answer for Perpendicular Lines - Download Free PDF

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3

Concept Used:

If two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

Calculation:

Given:

Line 1: \(\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}\)

Line 2: \(\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}\)

Lines are at right angles.

Line 1: \(\frac{x-1}{-3} = \frac{y-2}{\frac{2p}{7}} = \frac{z-3}{2}\)

Line 2: \(\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}\)

Direction ratios of Line 1: (-3, \(\frac{2p}{7}\), 2)

Direction ratios of Line 2: (\(\frac{-3p}{7}\), 1, -5)

Since the lines are perpendicular:

⇒ \((-3)\left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0\)

⇒ \(\frac{9p}{7} + \frac{2p}{7} - 10 = 0\)

⇒ \(\frac{11p}{7} = 10\)

 ⇒\(p = \frac{70}{11}\)

Hence option 3 is correct

Calculation

Given:

The lines p(p² + 1)x - y + q = 0 and (p² + 1)x + (p² + 1)y + 2q = 0 are perpendicular to a common line.

Then these lines must be parallel to each other.

∴ m₁ = m₂ ⇒ -\(\frac{p(p^2+1)}{-1}\) = -\(\frac{(p^2+1)^2}{p^2+1}\)

⇒ (p² + 1)(p + 1) = 0

⇒ p = -1

Hence option 1 is correct

Calculation

Direction ratios of line RQ:

(2-0, -3-(-1), -1-3) = (2, -2, -4) = (1, -1, -2)

Equation of line RQ in parametric form:

\(x = 0 + \lambda(1) = \lambda\)

\(y = -1 + \lambda(-1) = -1 - \lambda\)

\(z = 3 + \lambda(-2) = 3 - 2\lambda\)

Let the foot of the perpendicular from P on line RQ be F(\(\lambda\), -1-\(\lambda\), 3-2\(\lambda\)).

Direction ratios of PF:

(\(\lambda\)-1, -1-\(\lambda\)-8, 3-2\(\lambda\)-4) = (\(\lambda\)-1, -9-\(\lambda\), -1-2\(\lambda\))

Since PF is perpendicular to RQ, the dot product of their direction ratios is zero:

⇒ 1(\(\lambda\)-1) - 1(-9-\(\lambda\)) - 2(-1-2\(\lambda\)) = 0

⇒ \(\lambda\) - 1 + 9 + \(\lambda\) + 2 + 4\(\lambda\) = 0

⇒ 6\(\lambda\) + 10 = 0

⇒ 6\(\lambda\) = -10

⇒ \(\lambda\) = -10/6 = -5/3

Coordinates of F:

\(x = \lambda = -5/3\)

\(y = -1 - \lambda = -1 - (-5/3) = -1 + 5/3 = 2/3\)

\(z = 3 - 2\lambda = 3 - 2(-5/3) = 3 + 10/3 = 19/3\)

∴ The coordinates of the foot of the perpendicular are (-5/3, 2/3, 19/3).

Hence option 3 is correct

Concept:

Then invariant point of a transformation w = T(z) is given by z = T(z).

Explanation:

The invariant points of \(\rm w=\frac{2Z-4}{Z+2}\) is given by

\(\rm Z=\frac{2Z-4}{Z+2}\)

⇒ Z² + 2Z = 2Z - 4

⇒ Z² = -4

⇒ Z = ± 2i

(3) is true.

Concept:

Let two lines having direction ratios a_1, b₁, c_1, and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Given lines are  \(\frac{{2{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{{4 - {\rm{y}}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{2\left( {{\rm{x}} - 1} \right)}}{{2{\rm{k}}}} = \frac{{ - \left( {{\rm{y}} - 4} \right)}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}} \Leftrightarrow \frac{{\left( {{\rm{x}} - 1} \right)}}{{\rm{k}}} = \frac{{{\rm{y}} - 4}}{{ - 3}} = \frac{{{\rm{z}} + 2}}{{ - 1}}\)

So, the direction ratio of the first line is (k, -3, -1)

\(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\)

So, direction ratio of second line is (1, k, 4)

Lines are perpendicular,

∴ (k × 1) + (-3 × k) + (-1 × 4) = 0

⇒ k – 3k – 4 = 0

⇒ -2k – 4 = 0

∴ k = -2

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3

Concept:

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

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

2. Let two lines having direction ratio’s a_1, b₁, c₁ and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Note: Direction ratio’s of x-axis, y-axis and z-axis are (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively.

Calculation:

Given line is  \(\frac{{x - 1}}{1} = \frac{{y - 3}}{0} = \frac{{z - 2}}{5}\)

So, Direction ratio’s of the line is (1, 0, 5)

As we know that direction ratio’s of the y-axis is (0, 1, 0)

Now, apply the condition of perpendicular lines,

⇒ 1 × 0 + 0 × 1 + 5 × 0 = 0

Hence, y−axis and given line are perpendicular to each other.

Concept:

Let two lines having direction ratio’s a_1, b₁, c₁ and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Direction ratio’s of two lines are given as 1, 2, p − 1 and -3, 1, 2

Lines are perpendicular,

∴ 1 × -3 + 2 × 1 + (p – 1) × 2 = 0

⇒ -3 + 2 + 2p – 2 = 0

⇒ 2p = 3

∴ p = 3/2

Concept:

Let two lines having direction ratio’s a_1, b₁, c₁ and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Given point is P(1, 2, 3)

Let any point on x-axis is Q(k, 0, 0), where k is any real number

So direction ratio of PQ are (1 – k), (2 – 0), (3 – 0) = (1 – k), 2, 3

We know direction ratio of x-axis is given by 1, 0, 0

Now Line PQ perpendicular on x-axis

⇒ (1 – k) × 1 + 2 × 0 + 3 × 0 = 0

⇒ 1 – k = 0

∴ k = 1

Hence, the coordinate of a foot of perpendicular is (1, 0, 0)

Concept:

• If two lines are perpendicular to each other’s then product of direction ratio of these lines is zero.

Calculation:

Given lines are x = ay + b and z = cy + d

⇒ x – b = ay and z – d = cy

\(\therefore {\rm{\;}}\frac{{{\rm{x}} - {\rm{b}}}}{{\rm{a}}} = {\rm{\;}}\frac{{\rm{y}}}{1} = {\rm{\;}}\frac{{{\rm{z}} - {\rm{d}}}}{{\rm{c}}}\)            …. (1)

Direction ration of line is (a, 1, c)

Again, Lines are x = ey + f and z = gy + h

⇒ x – f = ey and z – h = gy

\(\therefore {\rm{\;}}\frac{{{\rm{x}} - {\rm{f}}}}{{\rm{e}}} = {\rm{\;}}\frac{{\rm{y}}}{1} = {\rm{\;}}\frac{{{\rm{z}} - {\rm{h}}}}{{\rm{g}}}\)            …. (2)

Direction ration of line is (e, 1, g)

Given both line are perpendicular

∴ (a × e) + (1 × 1) + (c × g) = 0

⇒ ae + 1 + cg = 0

Or, ae + cg + 1 = 0

Concept:

If the angle between the lines \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) where a1, b1, c1, a2, b2 and c2 are the direction ratios is 90° then

a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0

Calculation:

Given: Equation of two lines is \(\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2k}} = \frac{{z - 3}}{2}\;and\frac{{7 - 7x}}{{3k}} = \frac{{5 - y}}{1} = \frac{{6 - z}}{5}\)

The given equation can be re-written as:

\(⇒ \frac{{x - 1}}{{ - \;3}} = \frac{{y - 2}}{{\frac{{2k}}{7}}} = \frac{{z - 3}}{2}\;\ and \ \frac{{x - 1}}{{\frac{{ - \;3k}}{7}}} = \frac{{y - 5}}{{ - \;1}} = \frac{{z - 6}}{{ - \;5}}\)

The direction ratios of the given lines are: 

⇒ a1 = - 3, b1 = 2k/7, c1 = 2, a2 = - 3k/7, b2 = - 1 and c2 = - 5

∵ The given lines are perpendicular to each to each other

As we know that if two lines are perpendicular to each to each other then a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0

⇒ \({a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = \; - 3 \cdot \frac{{ - \;3k}}{7} + \left( {\frac{{2k}}{7}} \right) \cdot \left( { - \;1} \right) + 2 \cdot \left( { - \;5} \right) = 0\)

⇒ k = 10

Concept:

The coordinate axis divides the plane into four equal parts.

1. First Quadrant: (+, +)

2. Second Quadrant: (-, +)

3. Third Quadrant: (-, -)

4. Fourth Quadrant: (+, -)

Calculation:

According to the question, the side of the square is 3 units. Therefore,

There are two possible squares with one of the adjacent sides along the negative x-axis

Clearly, we can see that, possible coordinates of square are

Case 1: (0, 0), (-3, 0), (-3, 3) & (0, 3)

Case 1: (0, 0), (-3, 0), (-3, -3) & (0, -3)

Hence, coordinates (0, -3) & (3, 0) are not possible because point (3, 0) belongs to positive x-axis.

Hence, option 3 is correct.

Concept:

Let two lines having direction ratios a_1, b₁, c_1, and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Given lines are  \(\frac{{2{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{{4 - {\rm{y}}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{2\left( {{\rm{x}} - 1} \right)}}{{2{\rm{k}}}} = \frac{{ - \left( {{\rm{y}} - 4} \right)}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}} \Leftrightarrow \frac{{\left( {{\rm{x}} - 1} \right)}}{{\rm{k}}} = \frac{{{\rm{y}} - 4}}{{ - 3}} = \frac{{{\rm{z}} + 2}}{{ - 1}}\)

So, the direction ratio of the first line is (k, -3, -1)

\(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\)

So, direction ratio of second line is (1, k, 4)

Lines are perpendicular,

∴ (k × 1) + (-3 × k) + (-1 × 4) = 0

⇒ k – 3k – 4 = 0

⇒ -2k – 4 = 0

∴ k = -2

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3