Samacheer Class 11 Maths - Vector Algebra – I

ABC be the triangle and D, E, F is the midpoints of the sides AB, BC, and AC respectively.
Let O be the origin and let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are the position vectors of the points A, B and C respectively.
∴ \(\overrightarrow{\mathbf{D E}}\) is parallel to \(\overrightarrow{\mathbf{B C}}\) and length of DE is half the length of BC. Similarly we can prove \(\overrightarrow{\mathbf{E F}}\) = \(\frac{1}{2} \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{EF}}\)
parallel to \(\overrightarrow{\mathbf{A B}}\) and length of E is half the length of AB.
Also \(\overrightarrow{\mathbf{D F}}\) = \(\frac{1}{2} \overrightarrow{\mathrm{AC}}\)
\(\overrightarrow{\mathbf{D F}}\) is parallel to \(\overrightarrow{\mathbf{A C}}\) and length of DF is half the length of AC. Thus, the line segment joining the mid points of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.

Let ABCD be any quadrilateral D, E, F, G are the midpoints of the sides AB, BC, CD, AD respectively.
∴ EF and DG are parallel and equal.
∴ In the quadrilateral, DEFG’s opposite sides are parallel and equal. Therefore, DEFG is a parallelogram.

Let P, Q, R be the given points. Let O be the origin. Then the position vectors of
P, Q, R are \(\overrightarrow{\mathrm{OP}}\), \(\overrightarrow{\mathrm{OQ}}\) and \(\overrightarrow{\mathrm{OR}}\).
∴ P, Q, R lie on a straight line.
Hence, P, Q, R are collinear.

Let ABC be the triangle with centroid G. Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be the position vectors of the vertices A, B and C respectively with respect to the origin O. Then

Let ABC be the triangle. D, E, F are the mid points of the sides BC, CA, AB respectively. Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be the position vectors of the vertices A, B, C respectively. Then

(i) \(\frac{1}{5}, \frac{3}{5}, \frac{4}{5}\)
[If l, m, n are direction cosines of a vector then l 2 + m 2 + n 2 = 1]
∴ The given ratio \(\frac{1}{5}, \quad \frac{3}{5}, \quad \frac{4}{5}\) do not form the direction cosines of a vector.
(ii) \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\)
[If l, m, n are direction cosines of a vector then l 2 + m 2 + n 2 = 1]
∴ The given ratio form the direction cosines of a vector.
(iii) \(\frac{4}{3}, 0, \frac{3}{4}\)
[If l, m, n are direction cosines of a vector then l 2 + m 2 + n 2 = 1]
∴ The given ratio do not form the direction cosines of a vector.

(i) 1, 2, 3
The given direction ratios are a = 1, b = 2, c = 3
(ii) 3, – 1, 3
The given direction ratios are a = 3, b = – 1, c = 3
(iii) 0, 0, 7
The given direction ratios are a = 0, b = 0, c = 7

(ii) 3î + ĵ + k̂
(iii) ĵ
ĵ = 0î + ĵ + 0k̂
The direction ratios of the vector ĵ are (0, 1, 0)
The direction cosines of the vector ĵ are
(0, 1, 0)
Direction ratios = (0, 1, 0)
Direction cosines = (0, 1, 0)
(iv) 5î – 3ĵ – 48k̂
(v) 3î – 3k̂ + 4ĵ
(vi) î – k̂

Let ABC be the triangle and D, E, F is the midpoints of the sides BC, CA, AB respectively. Then AD, BE, CF are the medians of ∆ ABC.
Given that the vertices of the triangle are A (1, 0, 0), B (0, 1, 0 ) and C (0, 0, 1).

Given \(\frac{1}{2}, \frac{1}{\sqrt{2}}\), a are the direction cosines of some vector, then
[If l, m, n are direction cosines of a vector then l 2 + m 2 + n 2 = 1]

Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) \(\overrightarrow{\mathrm{OA}}=\hat{i}\) and \(\overrightarrow{\mathrm{OB}}=\hat{j}\)
Then \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\hat{j}-\hat{i}=-\hat{i}+\hat{j}\)
= (-1, 1, 0)
= (a, a + b, a + b + c)
⇒ a = -1, a + b = 1 and a + b + c = 0
Now a = -1 ⇒ -1 + b = 1;a + b + c = 0
⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0
⇒ c = -1
∴ a = -1; b = 2; c = -1.
Note: If we taken \(\overrightarrow{\mathrm{BA}}\) then we get a = 1, b = -2 and c = 1.

Let the given position vectors of the points A, B, C, D be
Equating the like terms on both sides
-4 = 3s – 7t ………… (1)
-6 = 10s – 5t ……….. (2)
-2 = 5s …………. (3)
(3) ⇒ s = \(-\frac{2}{5}\)
Substituting in equation (2), we have
-6 = 10 × \(-\frac{2}{5}\) – 5t
-6 = -4 – 5t
-6 + 4 = -5t
⇒ -5t = -2
⇒ t = \(\frac{2}{5}\)
Substituting for s and t in equation (1), we have

The given vectors are \(\vec{a}\) = 2î + 3ĵ – 4k̂, \(\vec{b}\) = 3î – 4ĵ – 5k̂ \(\vec{c}\) = – 3î + 2ĵ + 3k̂
(i) \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)
(ii) 3\(\vec{a}\) – 2\(\vec{b}\) + 5\(\vec{c}\)

Let A, B, C be the vertices of a triangle ABC.
Given the position vectors of the vertices A, B, C are given by
∴ The required perimeter of the triangle ABC is given by AB + BC + AC = \(\sqrt{44}+\sqrt{218}+\sqrt{110}\)

The given vector is m (î + ĵ + k̂)
Given that it is a unit vector.
∴ |m (i + j + k)| = 1
|m (i + j + k)| = 1
m (± \(\sqrt{1^{2}+1^{2}+1^{2}}\)) = 1
m = ± \(\frac{1}{\sqrt{3}}\)

The given vertices of a triangle are
A(1, 1, 1),B(1, 2, 3) and C (2, – 1, 1)
Position vector of A is \(\overrightarrow{\mathrm{OA}}\) = î + ĵ + k̂
Position vector of B is \(\overrightarrow{\mathrm{OB}}\) = î + 2ĵ + 3k̂
Position vector of C is \(\overrightarrow{\mathrm{OC}}\) = 2î – ĵ + k̂
∴ A, B, C are not collinear.
Hence A, B, C form a triangle.
AB = CA = √5
∴ ∆ABC is an isosceles triangle.

(i) \(\vec{a}\) = î – 2ĵ + k̂ and \(\vec{b}\) = 3î – 4ĵ – 2k̂
\(\vec{a}\). \(\vec{b}\) = (î – 2ĵ + k̂). (3î – 4ĵ – 2k̂)
= (1) (3) + (-2) (-4) + (1) (-2)
= 3 + 8 – 2
= 9
(ii) \(\vec{a}\) = 2î + 2ĵ – k̂ and \(\vec{b}\) = 6î – 3ĵ + 2k̂
\(\vec{a}\). \(\vec{b}\) = (2î + 2ĵ – k̂). (6î – 3ĵ + 2k̂)
= (2) (6) + (2) (-3) + (-1) (2)
= 12 – 6 – 2 = 12 – 8 = 4

When \(\vec{a}\) and \(\vec{b}\) are ⊥ r then \(\vec{a} \cdot \vec{b}\) = 0
\(\vec{a}\) ⊥ r \(\vec{b}\) ⇒ \(\vec{a} \cdot \vec{b}\) = 0
(i) (2) (1) + (λ) (-2) + (1) (3) = 0 ⇒ λ = 5/2
(ii) (2) (3) + (4) (-2) + (-1) (λ) = 0
6 – 8 – λ = 0
-λ – 2 = 0 ⇒ -λ = 2 ⇒ λ = -2

(i) 2î + 3ĵ – 6k̂ and 6î – 3ĵ + 2k̂
Let θ be the angle between the given vectors, then
(ii) î – ĵ and ĵ – k̂
Let θ be the angle between the given vectors, then

Given \(\vec{a}\) = 2î + 3ĵ + 6k̂ \(\vec{b}\) = 6î + 2ĵ – 3k̂ and \(\vec{c}\) = 3î – 6ĵ + 6k̂
\(\vec{a}\). \(\vec{b}\) = (2î + 3ĵ + 6k̂). (6î + 2ĵ – 3k̂)
= (2) (6) + (3) (2) + (6) (-3)
= 12 + 6 – 18
= 0
∴ \(\vec{a}\) and \(\vec{a}\) are perpendicular.
\(\vec{b}\). \(\vec{c}\) = (6î + 2ĵ – 3k̂). (3î – 6ĵ + 6k̂)
= (6) (3) + (2) (- 6) + (-3) (2)
= 18 – 12 – 6
= 0
∴ \(\vec{b}\) and \(\vec{c}\) are perpendicular.
\(\vec{c}\). \(\vec{a}\) = (3î – 6ĵ + 6k̂). (2î + 3ĵ + 6k̂)
= (3) (2) + (-6) (3) + (2) (6)
= 6 – 18 + 12
= 0
∴ \(\vec{c}\) and \(\vec{a}\) are perpendicular.
Hence \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors.

CA = \(\sqrt{1+9+25}\) = \(\sqrt{35}\)
AB ≠ BC + CA
∴ The given vectors form a triangle, Also
AB 2 = 41, BC 2 = 6, CA 2 = 35
AB 2 = BC 2 + CA 2
∴ ∆ ABC is a right angled triangle.

Let the given points be
A (2, -1, 3), B (4, 3, 1) and C (3, 1, 2)
AB = 2√6, BC = √6, CA = √6
BC + CA = √6 + √6 = 2√6
∴ BC + CA = BA = 2√6
Hence the given points A, B, C are collinear.

The given vectors are î + 3ĵ + 7k̂ and 2î + 6ĵ + 3k̂
Projection of î + 3ĵ + 7k̂ on 2î + 6ĵ + 3k̂ is

The given vectors are
\(\vec{a}\) = λî + ĵ + 4k̂, \(\vec{b}\) = 2î + 6ĵ + 3k̂
Also given that projection of \(\vec{a}\) and \(\vec{b}\) is 4 units.

(2) 60°
Explaination:
Given the angle made by \(\overrightarrow{\mathbf{O P}}\) with x – axis and y – axis are 60° and 45° respectively. Let the angle made by \(\overrightarrow{\mathbf{O P}}\) with the positive direction of z – axis be θ. Then

(3) cos -1 \(\left(\frac{1}{\sqrt{3}}\right)\)
Explaination:
Let the angles made by a vector with the coordinate axes be α, α, α. Then
cos 2 α + cos 2 α + cos 2 α = 1
[If α, β, γ are the angles made by a vector with coordinate axes respectively, then
cos 2 α + cos 2 β + cos 2 γ = 1]
3 cos 2 α = 1

(4) coplanar vectors
Explaination:
[The condition for the three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) to be coplanar is \(\vec{a}\) = λ\(\vec{a}\) + μ\(\vec{b}\) where λ, μ are scalars. That is one vector is a Linear combination of the other two vectors.]

(4) 8
Explaination:
The position vectors of the three points are
The condition for the three points A, B, C are collinear is the area of the triangle formed by these points is zero.