Find the vector equation of the plane in scalar product form r → = 2 i ^ + k ^ + λ i ^ + μ i ^ + 2 j ^ - 3 k ^

Given that r → = 2 i ^ + k ^ + λ i ^ + μ i ^ + 2 j ^ - 3 k ^ We know that, the equation r = a → + λ b → + μ c → represents a plane passing through a point whose position vector is a → . Here, a → = 2 i ^ + k ^ b → = i ^ c → = i ^ + 2 j ^ - 3 k ^ Normal vector, n → = b → × c → = i j k 1 0 0 1 2 - 3 = - j - 3 - 0 + k 2 - 0 = 3 j ^ + 2 k ^ The vector equation of the plane in scalar product form is r → . n → = a → . n → Therefore, r → . 3 j ^ + 2 k ^ = 2 .

Find the vector equation of the plane in scalar product form r → = 2 i ^ + k ^ + λ i ^ + μ i ^ + 2 j ^ - 3 k ^

Given that r → = 2 i ^ + k ^ + λ i ^ + μ i ^ + 2 j ^ - 3 k ^ We know that, the equation r = a → + λ b → + μ c → represents a plane passing through a point whose position vector is a → . Here, a → = 2 i ^ + k ^ b → = i ^ c → = i ^ + 2 j ^ - 3 k ^ Normal vector, n → = b → × c → = i j k 1 0 0 1 2 - 3 = - j - 3 - 0 + k 2 - 0 = 3 j ^ + 2 k ^ The vector equation of the plane in scalar product form is r → . n → = a → . n → Therefore, r → . 3 j ^ + 2 k ^ = 2 .

Show that the lines r → = 6 i ^ + j ^ + 2 k ^ + s i ^ + 2 j ^ - 3 k ^ and r → = 3 i ^ + 2 j ^ - 2 k ^ + t 2 i ^ + 4 j ^ - 5 k ^ are skew lines and hence find the shortest distance between them.

The vector equation of the given straight lines are: r → = 6 i ^ + j ^ + 2 k ^ + s i ^ + 2 j ^ - 3 k ^ . . . 1 r → = 3 i ^ + 2 j ^ - 2 k ^ + t 2 i ^ + 4 j ^ - 5 k ^ . . . 2 Compare these equations with the equation: r → = a → + s b → . . . 3 r → = c → + t d → . . . 4 From equations 1 and 3 a → = 6 i ^ + j ^ + 2 k ^ , b → = i ^ + 2 j ^ - 3 k ^ From equations 2 and 4 c → = 3 i ^ + 2 j ^ - 2 k ^ , d → = 2 i ^ + 4 j ^ - 5 k ^ The condition for the straight lines r → = a → + s b → and r → = c → + t d → to be skew lines is c → - a → . b → × d → ≠ 0 Now, c → - a → = 3 i ^ + 2 j ^ - 2 k ^ - 6 i ^ + j ^ + 2 k ^ ⇒ c → - a → = - 3 i ^ + j ^ - 4 k ^ And, b → × d → = i ^ j ^ k ^ 1 2 - 3 2 4 - 5 = 2 i ^ - j ^ So, b → × d → = 2 2 + - 1 2 = 5 So, c → - a → . b → × d → = - 3 i ^ + j ^ - 4 k ^ . 2 i ^ - j ^ = - 3 2 + 1 - 1 - 4 0 = - 6 - 1 - 0 = - 7 ≠ 0 So, the lines are skew lines. The shortest distance is given by: δ = c → - a → . b → × d → b → × d → = - 7 5 So, δ = 7 5 units .

Show that the lines x - 3 3 = y - 3 - 1 , z - 1 = 0 and x - 6 2 = z - 1 3 , y - 2 = 0 Intersect. Also find the point of intersection.

The equation of the given lines are x - 3 3 = y - 3 - 1 , z - 1 = 0 Let x - 3 3 = y - 3 - 1 = t ⇒ z - 1 = 0 ⇒ z - 1 0 = t So, the symmetric form of first line is x - 3 3 = y - 3 - 1 = z - 1 0 = t . . . 1 And, x - 6 2 = z - 1 3 , y - 2 = 0 Let x - 6 2 = z - 1 3 = s , y - 2 = 0 and y - 2 0 = s So, the symmetric form of second line is x - 6 2 = y - 2 0 = z - 1 3 = s . . . 2 From equations 1 and 2 x 1 , y 1 , z 1 = 3 , 3 , 1 x 2 , y 2 , z 2 = 6 , 2 , 1 b 1 , b 2 , b 3 = 3 , - 1 , 0 d 1 , d 2 , d 3 = 2 , 0 , 3 So, x 2 - x 1 y 2 - y 1 z 2 - z 1 b 1 b 2 b 3 d 1 d 2 d 3 = 6 - 3 2 - 3 1 - 1 3 - 1 0 2 0 3 = 3 - 1 0 3 - 1 0 2 0 3 = 3 - 3 - 0 + 1 9 - 0 + 1 0 + 2 = - 9 + 9 + 0 = 0 So, the given lines 1 and 2 intersect each other. So, symmetric form of second line is x - 6 2 = y - 2 0 = z - 1 3 = s . . . 2 Any point on line 1 is 3 t + 3 , - t + 3 , 0 t + 1 = 3 t + 3 , - t + 3 , 1 . . . 3 Any point on line 2 is 2 s + 6 , 0 s + 2 , 3 s + 1 = 2 s + 6 , 2 , 3 s + 1 . . . 4 Equations 3 and 4 represent the point of intersection of lines 1 and 2 if 3 t + 3 , - t + 3 , 1 = 2 s + 6 , 2 , 3 s + 1 For some values of s and t 3 t + 3 = 2 s + 6 ⇒ 3 t - 2 s = 3 And, - t + 3 = 2 ⇒ t = 1 And, 3 s + 1 = 1 ⇒ s = 0 So, the point of intersection of line 1 and 2 is obtained by putting t = 1 in 3 or s = 0 in 4 Put t = 1 in 3 , we get 3 × 1 + 3 , - 1 + 3 , 1 = 6 , 2 , 1 .

If the foot of the perpendicular drawn from the origin to the plane is ( 4 , - 2 , - 5 ) , then find the equation of the plane in vector and cartesian form.

To find the equation of the plane passing through the point ( 4 , - 2 , - 5 ) Whose position vector be a → = 4 i ^ - 2 j ^ - 5 k ^ and is normal to O P → i.e. n → = O P → = a → = 4 i ^ - 2 j ^ - 5 k ^ Thus vector equation of plane be r → · n → = a → · n → ⇒ r → · 4 i ^ - 2 j ^ - 5 k ^ = 4 i ^ - 2 j ^ - 5 k ^ · 4 i ^ - 2 j ^ - 5 k ^ ⇒ r → · 4 i ^ - 2 j ^ - 5 k ^ = 16 + 4 + 25 = 45 By putting r → = x i ^ + y j ^ + z k ^ we get x i ^ + y j ^ + z k ^ · 4 i ^ - 2 j ^ - 5 k ^ = 45 ⇒ 4 x - 2 y - 5 z = 45 is the required equation of plane.

Find the vector and Cartesian equations of the plane through the points ( 1 , 2 , 3 ) and ( 2 , 3 , 1 ) and perpendicular to the plane 3 x - 2 y + 4 z - 5 = 0

The required plane passes through the points A ( 1 , 2 , 3 ) and B ( 2 , 3 , 1 ) . It is perpendicular to the plane 3 x - 2 y + 4 z - 5 = 0 and therefore parallel to the normal vector v → = 3 i ^ - 2 j ^ + 4 k ^ The vector equation of the plane is of the form r → = 1 - s a → + s b → + t v → ⇒ r → = 1 - s i ^ + 2 j ^ + 3 k ^ + s 2 i ^ + 3 j ^ + k ^ + t 3 i ^ - 2 j ^ + 4 k ^ ⇒ r → = i ^ + 2 j ^ + 3 k ^ + s i ^ + j ^ - 2 k ^ + t 3 i ^ - 2 j ^ + 4 k ^ The Cartesian equation of the plane is of the form x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 l m n = 0 Where x 1 , y 1 , z 1 = ( 1 , 2 , 3 ) , x 2 , y 2 , z 2 = ( 2 , 3 , 1 ) , ( l , m , n ) = ( 3 , - 2 , 4 ) ⇒ x - 1 y - 2 z - 3 2 - 1 3 - 2 1 - 3 3 - 2 4 = 0 ⇒ x - 1 y - 2 z - 3 1 1 - 2 3 - 2 4 = 0 ⇒ ( x - 1 ) ( 4 - 4 ) - ( y - 2 ) ( 4 + 6 ) + ( z - 3 ) ( - 2 - 3 ) = 0 ⇒ - 10 y + 20 - 5 z + 15 = 0 or 2 y + z - 7 = 0 Hence, the vector equation of the plane is r → = i ^ + 2 j ^ + 3 k ^ + s i ^ + j ^ - 2 k ^ + t 3 i ^ - 2 j ^ + 4 k ^ and the Cartesian equation of the plane is 2 y + z - 7 = 0 .

Find the Cartesian equation of the plane containing the line x − 2 2 = y − 2 3 = z − 1 − 2 and passing through the point ( − 1 , 1 , − 1 ) .

The required plane contains the line x - 2 2 = y - 2 3 = z - 1 - 2 i.e., passes through ( 2 , 2 , 1 ) and is parallel to the vector 2 i → + 3 j → - 2 k → . Also it is passing through ( - 1 , 1 , - 1 ) x 1 , y 1 , z 1 = ( 2 , 2 , 1 ) ; x 2 , y 2 , z 2 = ( - 1 , 1 , - 1 ) ; l 1 , m 1 , n 1 = 2 , 3 , - 2 The required cartesian equation is, x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 l 1 m 1 n 1 = 0 ⇒ x - 2 y - 2 z - 1 - 3 - 1 - 2 2 3 - 2 = 0 ⇒ ( x - 2 ) ( 2 + 6 ) - ( y - 2 ) ( 6 + 4 ) + ( z - 1 ) ( - 9 + 2 ) = 0 ⇒ 8 ( x - 2 ) - 10 ( y - 2 ) - 7 ( z - 1 ) = 0 ⇒ 8 x - 16 - 10 y + 20 - 7 z + 7 = 0 ⇒ 8 x - 10 y - 7 z + 11 = 0

The vector equation of a plane passing through the line of intersection of the planes r → · n 1 → = q 1 and r → · n → 2 = q 2 is

r → · n → 1 - q 1 + λ r → · n 2 → - q 2 = 0

r → × n 1 → + r → × n 2 → = q 1 + q 2

r → × n 1 → - r → × n 2 → = q 1 + q 2

r → · n 1 → + r → · n 2 → = q 1 + λ q 2

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 19

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Earn 100

Find the vector equation of the plane in scalar product form $\vec{r} = (2 \hat{i} + \hat{k}) + λ (\hat{i}) + μ (\hat{i} + 2 \hat{j} - 3 \hat{k})$

Important Questions on Applications of Vector Algebra

MEDIUM

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 20

Find the vector of magnitude

9

which is equally inclined to the coordinate axes.

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 21

Find the non-parametric form of vector equation and Cartesian equation of the straight line passing through the point with position vector

4 \hat{i} + 3 \hat{j} - 7 \hat{k}

and parallel to the vector

2 \hat{i} - 6 \hat{j} + 7 \hat{k}

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 22

Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point

(- 2, 3, 4)

and parallel to the straight line

\frac{x - 1}{- 4} = \frac{y + 3}{5} = \frac{8 - z}{6}

.

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 23

Find the points where the straight line passes through

(6, 7, 4)

and

(8, 4, 9)

cuts the

x z

and

y z

planes.

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 24

If the straight line joining the points

(2, 1, 4) and (a - 1, 4, - 1)

is parallel to the line joining the points

(0, 2, b - 1)

and

(5, 3, - 2)

, find the values of

a and b .

HARD

12th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 6 - Applications of Vector Algebra>Exercise>Q 25

Find the parametric form of vector equation and Cartesian equations of a straight line passing through $(5, 2, 8)$ and is perpendicular to the straight lines

$\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + s (2 \hat{i} - 2 \hat{j} + \hat{k})$ and