Find the equation of plane containing the two lines x - 1 2 = y + 1 - 1 = z 3 and x - 2 = y - 2 - 3 = z + 1 - 1 .

Two given lines are x - 1 2 = y + 1 - 1 = z - 0 3 and x - 0 - 2 = y - 2 - 3 = z + 1 - 1 Now, if these two lines lie on a plane and if the normal vector of plane is n → = a i ^ + b j ^ + c k ^ , then both lines must be perpendicular to n → , hence 2 a - b + 3 c = 0 - 2 a - 3 b - c = 0 ⇒ a 1 + 9 = - b - 2 + 6 = c - 6 - 2 = k ⇒ a = 10 k , b = - 4 k , c = - 8 k Also, plane is passing through the point 1 , - 1 , 0 . Hence, equation of plane is a x - 1 + b y + 1 + c z - 0 = 0 ⇒ 10 x - 1 - 4 y + 1 - 8 z - 0 = 0 ⇒ 5 x - 1 - 2 y + 1 - 4 z = 0 ⇒ 5 x - 2 y - 4 z = 7

If the points 1 , 1 , p and - 3 , 0 , 1 be equidistant from the plane r → · 3 i ^ + 4 j ^ - 12 k ^ + 13 = 0 , then find the value of p .

The given points are 1 , 1 , p and - 3 , 0 , 1 The position vector of the point 1 , 1 , p is a → 1 = i ^ + j ^ + p k ^ Similarly, the position vector of the point - 3 , 0 , 1 is a → 2 = - 3 i ^ + k ^ The equation of the given plane is r → · 3 i ^ + 4 j ^ - 12 k ^ + 13 = 0 We know that, the perpendicular distance from a point whose position vector is a → to the plane, r → · N → = d , is given by D = a · → N → - d N → In this condition, N → = 3 i ^ + 4 j ^ - 12 k ^ and d = - 13 Thus, the distance from the point 1 , 1 , p to the given plane is D 1 = i ^ + j ^ + p k ^ · 3 i ^ + 4 j ^ - 12 k ^ + 13 3 i ^ + 4 j ^ - 12 k ^ ⇒ D 1 = 3 + 4 - 12 p + 13 3 2 + 4 2 + ( - 12 ) 2 ⇒ D 1 = | 20 - 12 p | 13 …(i)| Distance from the point - 3 , 0 , 1 to the given plane is D 2 = - 3 i ^ + k ^ · 3 i ^ + 4 j ^ - 12 k ^ + 13 3 i ^ + 4 j ^ - 12 k ^ ⇒ D 2 = | - 9 - 12 + 13 | 3 2 + 4 2 + ( - 12 ) 2 ⇒ D 2 = 8 13 …(ii) Since, it is given that the distance to the plane from the points, 1 , 1 , p and - 3 , 0 , 1 is equal. ∴ D 1 = D 2 ⇒ | 20 - 12 p | 13 = 8 13 ⇒ 20 - 12 p = 8 or - ( 20 - 12 p ) = 8 ⇒ 12 p = 12 or 12 p = 28 ⇒ p = 1 or p = 7 3 Hence, the value of p is 1 or 7 3 .

i Find the point where the plane r → · i ^ + 2 j ^ + k ^ = 4 meets the line r → = 2 i ^ + j ^ - k ^ + t i ^ + j ^ + 2 k ^ . ii Find the angle that the line makes with the plane.

We have, r → = 2 i ^ + j ^ - k ^ + t i ^ + j ^ + 2 k ^ ⇒ x - 2 1 = y - 1 1 = z + 1 2 = k . . . . i Any point in this line is P k + 2 , k + 1 , 2 k - 1 . Let the required point be P . r → · i ^ + 2 j ^ + k ^ = 4 ⇒ x + 2 y + z = 4 . . . . ii If a line is meeting the plane at P , then k + 2 + 2 k + 1 + 2 k - 1 = 4 ⇒ 5 k = 1 ⇒ k = 1 5 Hence, line is meeting the plane at P 11 5 , 6 5 , - 3 5 . ii Angle between line and plane is given by = sin - 1 1 + 2 + 2 6 × 6 = sin - 1 5 6 .

Show that the lines x - a + d α - δ = y - a α = z - a - d α + δ and x - b + c β - γ = y - b β = z - b - c β + γ are coplanar.

Consider two points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) and two vectors a → and b → having direction ratios as ( a 1 , b 1 , c 1 ) and ( a 2 , b 2 , c 2 ) respectively. Then the condition for them to be coplanar is A B → . ( a → × b → ) = 0 . The two lines given in the question are x - a + d α - δ = y - a α = z - a - d α + δ and x - b + c β - γ = y - b β = z - b - c β + γ . These can be related to the form x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 . So, the points for first line are x 1 = a - d , y 1 = a , z 1 = a + d , a 1 = α - δ , b 1 = α , c 1 = α + δ and similarly, for the second line, they are x 2 = b - c , y 2 = b , z 2 = b + c , a 2 = β - γ , b 2 = β , c 2 = β + γ . Now the direction ratios of the vector A B → = b - c - a - d , b - a , b + c - a + d ⇒ b - c - a + d , b - a , b + c - a - d . Now applying the condition for coplanarity x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = b - c - a + d b - a b + c - a - d α - δ α α + δ β - γ β β + γ = 2 ( b - a ) b - a b + c - a - d 2 α α α + δ 2 β β β + γ C 1 → C 1 + C 3 = 2 3 ( b - a ) b - a b + c - a - d α α α + δ β β β + γ = 2 3 0 b - a b + c - a - d 0 α α + δ 0 β β + γ = 0 Since the condition is satisfied, both the lines are coplanar.

Find the equations of straight line perpendicular to the lines x + 1 - 3 = y - 3 2 = z + 2 1 and x 1 = y - 7 - 3 = z + 7 2 , and passing through their point of intersection.

Given two lines x + 1 - 3 = y - 3 2 = z + 2 1 and x 1 = y - 7 - 3 = z + 7 2 . Let x + 1 - 3 = y - 3 2 = z + 2 1 = r So any point on the first line is - 3 r - 1 , 2 r + 3 , r - 2 . . . i It is the point of intersection of the lines if the point lies on the second line, then - 1 - 3 r 1 = 3 + 2 r - 7 - 3 = - 2 + r + 7 2 ⇒ - 1 - 3 r 1 = 3 + 2 r - 7 - 3 ⇒ 9 r + 3 = 2 r - 4 ⇒ r = - 1 This value of r satisfies 3 + 2 r - 7 - 3 = - 2 + r + 7 2 because 2 - 1 - 4 - 3 = - 1 + 5 2 or - 6 - 3 = 4 2 , which is true So, using equation i , the point of intersection is 2 , 1 , - 3 ∴ The required line is x - 2 l = y - 1 m = z + 3 n . . . ii This line is perpendicular to both given lines, So using condition of perpendicularity - 3 l + 2 m + n = 0 l - 3 m + 2 n = 0 ∴ l 4 - - 3 = m 1 - - 3 2 = n - 3 - 3 - 2 ⇒ l 7 = m 7 = n 7 = k (suppose). ∴ Using equation ii , the required line is x - 2 7 k = y - 1 7 k = z + 3 7 k or x - 2 1 = y - 1 1 = z + 3 1

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

The equation of the given lines are: x - 5 7 = y + 2 - 5 = z 1 & x 1 = y 2 = z 3 The direction ratios of these lines are 7 , - 5 , 1 & 1 , 2 , 3 . Two lines with direction ratios a 1 , b 1 , c 1 and a 2 , b 2 , c 2 are perpendicular to each other if a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 ∴ a 1 a 2 + b 1 b 2 + c 1 c 2 = ( 7 × 1 ) + ( - 5 × 2 ) + ( 1 × 3 ) = 7 - 10 + 3 = 10 - 10 = 0 Thus, given lines are perpendicular to each other.

Find the distance between the lines ℓ 1 and ℓ 2 given by r → = i ^ + 2 j ^ - 4 k ^ + λ 2 i ^ + 3 j ^ + 6 k ^ and r → = 3 i ^ + 3 j ^ - 5 k ^ + μ 2 i ^ + 3 j ^ + 6 k ^ .

The two lines are parallel. We have, a → 1 = i ^ + 2 j ^ - 4 k ^ , a → 2 = 3 i ^ + 3 j ^ - 5 k ^ and b → = 2 i ^ + 3 j ^ + 6 k ^ . Therefore, the distance between the lines is given by d = b → × a → 2 - a → 1 | b → | b → × a → 2 - a → 1 = i ^ j ^ k ^ 2 3 6 2 1 - 1 = - 9 i ^ + 14 j ^ - 4 k ^ b → = 4 + 9 + 36 = 49 d = | - 9 i ^ + 14 j ^ - 4 k ^ | 49 = 293 49 = 293 7 units

Find the angle between the pair of lines r → = 3 i ^ + 2 j ^ - 4 k ^ + λ i ^ + 2 j ^ + 2 k ^ and r → = 5 i ^ - 2 j ^ + μ 3 i ^ + 2 j ^ + 6 k ^ .

Given, the pair of lines is r → = 3 i ^ + 2 j ^ - 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) and r → = 5 i ^ - 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) a 1 → = 3 i ^ + 2 j ^ - 4 k ^ and a 2 → = 5 i ^ - 2 j ^ b 1 → = i ^ + 2 j ^ + 2 k ^ and b 2 → = 3 i ^ + 2 j ^ + 6 k ^ Now, cos θ = b 1 → · b 2 → | b 1 → | | b 2 → | ⇒ cos θ = ( i ^ + 2 j ^ + 2 k ^ ) · ( 3 i ^ + 2 j ^ + 6 k ^ ) 1 2 + 2 2 + 2 2 3 2 + 2 2 + 6 2 ⇒ cos θ = 3 + 4 + 12 3 7 ⇒ cos θ = 19 21 ∴ θ = cos - 1 19 21 .

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 27

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Find the foot of the perpendicular drawn from the point $(6, - 2, 5)$ to the straight line passing through the point $(4, - 1, 2)$ having direction ratios $1, 4, - 3$ .

Important Questions on Three Dimensional Geometry

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 28

A line

\frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z - \frac{9}{2}}{3}

cuts the

y - z

plane and

x - y

plane at

A

and

B

respectively. Find

∠ A O B, O

being origin.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 29

Find the equation of plane containing the two lines

\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{3}

and

\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}

.

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 30

Find the equation of the plane which passes through the point

(3, 2, 0)

and contains the line

\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}

.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 31

Find the equation of the plane through the intersection of the planes

3 x - y + 2 z - 4 = 0

and

x + y + z - 2 = 0

and the point

(2, 2, 1)

.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 32

If the points

(1, 1, p)

and

(- 3, 0, 1)

be equidistant from the plane

\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,

then find the value of

p .

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 33

$(i)$ Find the point where the plane $\vec{r} \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = 4$ meets the line $\vec{r} = (2 \hat{i} + \hat{j} - \hat{k}) + t (\hat{i} + \hat{j} + 2 \hat{k})$ .

$(ii)$ Find the angle that the line makes with the plane.

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 32 - Three Dimensional Geometry>Assignment>Q 34

Show that the lines $\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}$ and $\frac{x - b + c}{β - γ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}$ are coplanar.