Vector Equation of Line Through Two Points Class 12 OP Malhotra Exe-23C ISC Maths Solutions Ch-23 Three Dimensional Geometry. In this article you would learn How to find a Vector equation of a line passing through two given points. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Vector Equation of Line Through Two Points Class 12 OP Malhotra Exe-23C ISC Maths Solutions Ch-23 Three Dimensional Geometry

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-23	Three Dimensional Geometry
Writer	OP Malhotra
Exe-23(c)	Vector Equation of Line Through Two Points

Vector Equation of Line Through Two Points

Ch-23 Three Dimensional Geometry Class 12 OP Malhotra Exe-23C Solutions

Que-1: Find the vector equation of a line which passes through the vector 2 î + 4 ĵ + 5k̂ and is parallel to the vector 2 î + 2 ĵ – 3k̂. Find the Cartesian form also.

Sol: We know that the vector equation of line passing through the point whose position vector is a→ and || to vector b→ is r→= a→ + λb→ , where λ be the some scalar.
Thus required vector eqn. of line through
a→ = 3 î + 4ĵ + 5k̂ and parallel to b→ = 2î + 2ĵ – 3k̂ is given by
r→ = 3 î + 4ĵ + 5k̂ + λ(2î + 2ĵ – 3k̂)
Let r→ = x î + y ĵ + z k^ be the P.V. of any point on line (1).
∴ eqn. (1) becomes; x î + yĵ + zk̂
= (3 + 2λ) î + (4 + 2λ)ĵ+ (5 – 3λ)k̂
Comparing the coefficients of î, ĵ, k̂ on both sides, we have
x = 3 + 2λ
⇒ x−3/2 = λ
y = 4 + 2λ
⇒ y−4/2 = λ
z = 5 – 5λ
⇒ (z−5)/−3 = λ
Thus x−3/2 = y−4/2 = (z−5)/−3 be the required eqn. of line in cartesian form.

Que-2: Find the vector equation of a line which is parallel to the vector 2î – ĵ + 3k̂ and which passes through the point (5, -2, 4). Reduce it to the Cartesian form.

Sol: We know that vector equation of line passes through the point with position vector a⃗ and parallel to vector b→ is given by
r→ = a→ + λ b→
Here a→ = 5î – 2ĵ + 4k̂
and b→ = 2î – ĵ + 3 k̂
Thus the required vector eqn. of line be r→ = 5î – 2ĵ + 4k̂ + λ(2î –ĵ + 3k̂)
Let r→ = xî + yĵ  + zk̂ be the position vector of any point on this line
∴ From (1) we get xî + yĵ + zk̂
= 5î – 2ĵ + 4k̂ + λ(2î – ĵ + 3k̂)
⇒ xî + yĵ + zk̂
= (5 + 2 λ)î+ (-2 – λ) ĵ + (4 + 3λ)k̂
On equating the coefficients of î,ĵ and k̂ on both sides
we have x = 5 + 2λ, y = -2 – λ and z = 4 + 3λ
⇒ x−5/2 = λ; (y+2)/−1 = λ;
z−43 = λ
i.e. x−5/2 = (y+2)/−1 = z−4/3
[on eliminating λ] be the required eqn. of line in cartesian form.

Que-3: Find the vector and Cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).

Sol: Let a→ and b→ be the position vectors of the given points A(3, -2, -5) and B(3, -2, 6)
Thus required vector eqn. of line through the points A and B with P.V. a→ and b→ is given by
r→ = a→ + λ(b→ – a→ )
∴ r→ = 3î – 2ĵ – 5k̂ +
λ[(3 î – 2ĵ + 6k̂) – (3î – 2ĵ – 5k̂)]
⇒ r→ = 3î – 2ĵ – 5k̂ + λ(0î + 0ĵ + 11k̂)
⇒ r→ = 3î – 2ĵ + (11λ – 5)k̂
The cartesian eqn. of line be
x−3/3−3 = y(−2)/−2+2 = z−(−5)/6−(−5)
i.e., x−3/0 = y+2/0 = z+5/11

Que-4: Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î – ĵ + 4k̂ and is in the direction î + 2 ĵ – k̂

Sol: We know that vector eqn. of line passing through the point whose P.V. a→ and || to vector b→ is
r→ = a→ + λb→
Thus required vector eqn. of line through the point with P.V. 2î – ĵ + 4 k̂ and in the direction of î + 2 ĵ – k̂ is
r→ = 2î – ĵ + 4k̂ + λ(î + 2ĵ – k̂)
Let r→ = xî + yĵ + zk̂  be the P.V. of any point on line (1)
∴ eqn. (1) becomes; xî + yĵ + zk̂ = (2 + λ)î + (-1 + 2λ)ĵ+ (4-λ)k̂
On comparing the coefficients of î, ĵ and k^ on both sides, we have
x = 2 + λ
⇒ x−2/1 = λ;
z = 4 – λ
⇒ z−4/−1 = λ
y = -1 + 2λ
⇒ y+1/2 = λ
Thus, x−2/1 = y+1/2 = z−4/−1 be the required equal line in cartesian form.

Que-5: The Cartesian equations of a line are 6x – 2 = 3y + 1 = 2z – 2. Find the vector equation of this line.

Sol: Eqn. of given line be,
6x – 2 = 3y + 1 = 2z – 2
⇒ 6(x – (1/3)) = 3(y + (1/3)) = 2(z – 1)
⇒ (x−1/3)/(1/6) = (y+(1/3))/(1/3) = z−1/(1/2)
⇒ (x−(1/3))/1 = (y+(1/3))/2 = z−1/3
∴ D’ratios of given line are proportional to < 1, 2, 3 >.
Since the required line is || to given line
∴ required line || to the vector b⃗ = î + 2ĵ + 3k̂
Hence the vector eqn. of line which passing through the point (1/3, −1/3, 1) and || to vector b⃗ is
r⃗ = 13î – 13ĵ + k̂ + λ(î + 2ĵ + 3k̂ )

Que-6: Find the vector equation of a line passing through the point with position vector î – 2ĵ – 3k̂ and parallel to the line joining the points with position vectors î – ĵ + 4k̂ and 2î + ĵ + 2k̂ . Also find the Cartesian form of the equation.

Sol: Let A, B, C be the given points whose position vectors are
î – 2ĵ – 3k̂ ; î – ĵ + 4k̂ ; 2î + ĵ + 2k̂
we want to find the eqn. of line passing through A and parallel to BC→
Now BC→ = P.V. of C – P. V. of B = (2î + ĵ + 2k̂) – (î – ĵ + 4k̂) = î + 2ĵ – 2k̂
also vector eqn. of line passes through point A with P.V. a→ and || to b→ is given by r→ = a→ + λb→
Here a→ = î – 2 ĵ – 3k̂ and b→ = î + 2 ĵ – 2k̂
i.e. r→ = î – 2ĵ – 3k̂ + λ(î + 2 j^ – 2k̂) be the required vector equation.
putting r→ = x î + y ĵ + zk̂ in this line, we get
xî + yĵ + zk̂ = (1 + λ)î + (-2 + 2λ) ĵ + (-3 – 2)k̂
equating the coefficients of î, ĵ, k̂ on both sides, we get
∴x = 1 + λ ; y = -2 + 2 λ ; z = -3 – 2 λ
⇒ x−1/1 = λ ; y+2/2 = λ; z+3/−2 = λ
On eliminating λ we get ; x−1/1 = y+2/2 = z+3/−2 be the required cartesian equation of line.

Que-7: Show that the points, whose position vectors are given by
-4î + 2ĵ – 3k̂, î + 3ĵ – 2k̂ and (-9î + ĵ – 4k̂) are collinear.

Sol: Let A, B, C be the points with a→ , b→ , c→ be then refrective position vectors.
Then a→ = -4î + 2ĵ – 3k̂ ; b→ = î + 3ĵ – 2k̂ and c→ = -9î +ĵ – 4k̂
∴ vector eqn. of line passing through the points A(a⃗ ) and B(b→ ) is given by
r→ = a→ + λ(b→ – a→ )
⇒r→ = (-4î + 2ĵ – 3k̂) + λ[(î + 3ĵ – 2k̂) – (-4î + 2ĵ – 3k̂)]
⇒ r→ = (-4î + 2ĵ – 3k̂) + λ(5î + ĵ + k̂)
Now the given points A, B and C are collinear if point C lies on eqn. (1).
i.e. if point C lies on eqn. (1).
i.e., if P.V. of c→ i.e. -9î + ĵ – 4k̂ satisfies eqn. (1).
i.e., if -9î +ĵ – 4k̂ = (-4î + 2ĵ – 3k̂) + λ(5î + ĵ + k̂)
i.e., if -9 = -4 + 5λ, 1 = 2 + λ, -4 = -3 + λ i.e., if λ = -1, λ = -1, λ = -1
Hence, λ = 1 satisfies all equation. Thus the given points are collinears.

Que-8: The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, -1) are three vertices of a parallelogram A B C D. Find the vector and Cartesian equations for the sides A B and B C and find the coordinates of D.

Sol: The vector eqn, of line through the point A(a⃗ ) and B(b⃗ ) is given by
r→ = a→ + λ(b→ – a→ )
where a→ = 4î + 5ĵ + 10k̂, b→
= 2î + 3ĵ + 4k̂ and c→ = î + 2ĵ – k^ Hence, vector eqn. of side A B is given by
r→ = 4î + 5ĵ + 10k̂ + λ[(2î + 3ĵ + 4k̂) – (4î + 5ĵ + 10k̂)]
i.e. r→ = 4î + 5ĵ + 10k̂ + λ(-2î – 2ĵ – 6k̂)
and cartesian equ. of AB be given by
x−4/2−4 = y−5/3−5 = z−10/4−10
i.e., x−4/−2 = y−5/−2 = z−10/−6
i.e., x−4/1 = y−5/1 = z−10/3
Therefore, vector eqn. of line BC be given by
r→ = b→ + A(c→ – b→ )
i.e., r→ = 2î + 3ĵ + 4k̂ + λ(î + 2ĵ –k̂) – (2î + 3ĵ + 4k̂)]
i.e., r→ = 2î + 3ĵ + 4k̂ + λ(−î – ĵ – 5k̂)
i.e., r→ = 2î + 3ĵ + 4k̂ + λ(î +ĵ + 5k̂);
where λ = -λ
Hence, required cartesian equation of line C D is given by
x−2/1−2 = y−3/2−3 = z−4/−1−4
i.e., x−2/−1 = y−3/−1 = z−4/−5
i.e., x−2/1 = y−3/1 = z−4/5
LetD(α, β, γ) be the coordinates of vertex D of || gm ABCD. Since diagonals AC and BD bisect each other.
∴ Mid point of AC = mid point of BD
i.e., (4+1/2, 5+2/2, 10−1/2)
= (2+α/2, 3+β/2, 4+γ/2)
i.e., 5/2 = 2+α/2
⇒ α = 3 ; 7/2 = 3+β/2
⇒ β = 4 ; 9/2 = 4+γ/2
⇒ γ = 5
Thus, required coordinates of D are (3, 4, 5).

Que-9: Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line −x−2/1 = y+3/7 = 2z−6/3

Sol: eqn. of given line be −x−2/1 = y+3/7 = 2z−6/3
i.e. x+2/−1 = y+3/7 = z−3/(3/2)
∴ D’ratios of given line (1) proportional to < -1, 7, > 3/2
i.e. < -2, 14, 3 >
Thus Direction ratios of required line i.e. line || to line (1) are proportional to < -2, 14, 3 >
Hence the catesian eqn. of line passing through (1, 2, 3) having direction ratios < -2, 14, 3 > is given by x−1/−2 = y−2/14 = z−3/3 = t (say)
⇒ x = -2 t + 1 ; y = 14 t + 2 ; z = 3 t + 3
Let r→ = xî + yĵ + z k^ be P.V. of any point on line
Then r→ = (-2 t + 1)î + (14 t + 2)ĵ + (3 t + 3)k̂
⇒ r→ = î + 2ĵ + 3k̂ + t(-2î + 14ĵ + 3k̂) be the required vector eqn. of line.

Que-10: Find the direction cosines and vector equation of the line whose Cartesian form is x−2/2 = 2y−5/−3, z = -1.

Sol: Given eqn. of line in cartesian form be
x−2/2 = 2y−5/−3 = z+1/0
i.e., x−2/2 = y−5/2/−(3/2) = z+1/0
∴ eqn. of given lines be passes through the point (2, 5/2, -1)
and having direction ratios < 2, −3/2, 0 >
i.e., < 4, -3, 0 >
Thus, vector eqn. of line passes through the point (2, 5/2, -1) whose P.V. be a→ = 2î + 5/2ĵ – k̂ and || to vector b→ = 4î -3ĵ + 0k̂ is given by
r→ = a→ + λb→
i.e., r→ = 2î + 5/2 ĵ – k̂ + λ(4î – 3ĵ + 0k̂)
∴ direction cosines of line be < 4/√4²+(−3)²+0², −3/√4²+(−3)²+0², 0 >
i.e., < 4/5, −3/5, 0 >

–: End of Vector Equation of Line Through Two Points Class 12 OP Malhotra Exe-23C ISC Maths Solutions :–

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