Revision: Mathematics >> Vectors CUET (UG) Vectors

Vectors that are perpendicular to each other are called orthogonal vectors.

Two vectors having the same magnitude and the same direction are called equal vectors.

A vector that describes the position of a point with respect to the origin is called a position vector.

A vector that has a magnitude of one unit and is used to indicate direction is called a unit vector.

A physical quantity that is described with both magnitude and direction is called a vector.

A vector whose magnitude is zero is called a zero vector.

A vector that has the same magnitude as a given vector but acts in the opposite direction is called a negative vector.

A physical quantity that is described with magnitude alone is called a scalar.

When a vector \[\vec P\] is split into two mutually perpendicular parts along the horizontal and vertical axes, those parts are called rectangular components.

Vectors that act in the same plane are called coplanar vectors.

The vectors which act in the same plane are called co-planar vectors.

A vector having the same magnitude as the original vector but having an opposite direction is called the negative of a vector.

The length or the magnitude of a vector is called the modulus of a vector.

A vector of unit magnitude drawn in the direction of a given vector is called a unit vector.

A vector that has zero magnitude and an arbitrary direction, represented by \[\vec 0\], is called a zero vector or null vector.

A vector is any quantity that needs both magnitude (size) and direction to be completely described.

OR

The physical quantities which have both magnitude and direction, obey the laws of vector addition, and are specified by a number with a unit and its direction (e.g., displacement, velocity, force, momentum) are called vector quantities or vectors.

If ā and b̄ are any two vectors, then the scalar product of these vectors is

ā · b̄ = |ā| |b̄| cos θ = ab cos θ, where θ is the angle between ā and b̄.

The scalar triple product of three vectors a, b, and c is defined as

(a × b) · c = |a| |b| |c| sinθ cosφ,

where θ is the angle between a and b, and φ is the angle between a × b and c. It is also defined as [a b c].

\[\vec P\] ⋅ \[\vec Q\] ​= PQ cos θ

If two vectors \[\vec P\] and \[\vec Q\]​ act at an angle θ, the magnitude of their resultant is:

If a vector \[\vec P\] is resolved into two rectangular components:

• Horizontal component: P_x = P cos⁡ θ
• Vertical component: P_y = P sin ⁡θ

∣\[\vec P\] × \[\vec Q\]​∣ = PQ sin θ

\[P\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\]

\[\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\]

Parallelepiped: Volume = [a b c]

Tetrahedron: \[\frac{1}{6}\] [a b c]

If two vectors are represented in magnitude and direction by two sides of a triangle taken in order, then their sum is \[\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}\]

The resultant R of two vectors \[\vec P\] and \[\vec Q\]​ always lies between their difference and their sum:

For three vectors \[\vec P\], \[\vec Q\]​, and \[\vec R\]:

The associative law holds true for addition of vectors but not for subtraction.

For any two vectors \[\vec P\] and \[\vec Q\]​:

The commutative law holds true for addition of vectors but not for subtraction.

Two vectors can be added using either the Triangle Law or the Parallelogram Law. When two vectors \[\vec P\] and \[\vec Q\]​ are represented as two sides of a triangle (or two adjacent sides of a parallelogram), their resultant is represented by the third side (or the diagonal).

If two co-initial vectors are represented in magnitude and direction by adjacent sides of a parallelogram, then their sum is represented in magnitude and direction by the diagonal of the parallelogram passing through the common point.

∴ \[\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}\]

If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in the same order, then their resultant is represented both in magnitude and direction by the closing side of the polygon taken in opposite order — this is called the Polygon Law of Vector Addition.

If two vectors can be represented both in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented both in magnitude and direction by the third side of the triangle taken in the opposite order — this is called the Triangle Law of Vector Addition.

If two vectors can be represented both in magnitude and direction by the two adjacent sides of a parallelogram drawn from a common point, then their resultant is completely represented, both in magnitude and direction, by the diagonal of the parallelogram passing through that point — this is called the Parallelogram Law of Vector Addition.

R is a point on the line segment AB(A – R – B) and `bar(AR)` and `bar(RB)` are in the same direction.

Point R divides AB internally in the ratio m : n

∴ `(AR)/(RB) = m/n`

∴ n(AR) = m(RB)

As `n(bar(AR))` and `m(bar(RB))` have same direction and magnitude,

`n(bar(AR)) = m(bar(RB))`

∴ `n(bar(OR) - bar(OA)) = m(bar(OB) - bar(OR))`

∴ `n(vecr - veca) = m(vecb - vecr)`

∴ `nvecr - nveca = mvecb - mvecr`

∴ `mvecr + nvecr = mvecb + nveca`

∴ `(m + n)vecr = mvecb + nveca`

∴ `vecr = (mvecb + nveca)/(m + n)`

As the point R divides the line segment AB externally, we have either A-B-R or R-A-B.

Assume that A-B-R and `bar(AR) : bar(BR)` = m : n

∴ `(AR)/(BR) = m/n` so n(AR) = m(BR) 

As `n(bar(AR))` and `m(bar(BR))` have same magnitude and direction,

∴ `n(bar(AR)) = m(bar(BR))`

∴ `n(barr - bara) = m(barr - barb)`

∴ `nbarr - nbara = mbarr - mbarb`

∴ `mbarr - nbarr = mbarb - nbara`

∴ `(m - n)barr = mbarb - nbara`

∴ `barr = (mbarb - nbara)/(m - n)`

Hence proved.

Let `bara, barb, barc, bard, bare, barf` be the position vectors of the points A, B, C, D, E, F respectively.

Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula

`bard = (barb + barc)/2, bare = (barc + bara)/2, barf = (bara + barb)/2`

∴ `bar(AD) + bar(BE) + bar(CF) = (bard - bara) + (bare - barb) + (barf - barc)`

= `((barb + barc)/2 - bara) + ((barc + bara)/2 - barb) + ((bara + barb)/2 - barc)`

= `1/2barb + 1/2barc - bara + 1/2barc + 1/2bara - barb + 1/2bara + 1/2barb - barc`

= `1/2(barb + barc - 2bara + bar c + bara - 2barb + bara + barb - 2barc)`

= `(bara + barb + barc) - (bara + barb + barc) = bar0`. 

Let A, B and C be vertices of a triangle.

Let D, E and F be the mid-points of the sides BC, AC and AB respectively.

Let `bara, barb, barc, bard, bare` and `barf` be position vectors of points A, B, C, D, E and F respectively.

Therefore, by mid-point formula,

∴ `bard = (barb + barc)/2, bare = (bara + barc)/2` and `barf = (bara + barb)/2`

∴ `2bard = barb + barc, 2bare = bara + barc` and `2barf = bara + barb`

∴ `2bard + bara = bara + barb + barc`, similarly `2bare + barb = 2barf + barc = bara + barb + barc`

∴ `(2bard + bara)/3 = (2bare + barb)/3 = (2barf + barc)/3 = (bara + barb + barc)/3 = barg`  ...(Say)

Then we have `barg = (bara + barb + barc)/3 = ((2)bard + (1)bara)/(2 + 1) = ((2)bare + (1)barb)/(2 + 1) = ((2)barf + (1)barc)/(2 + 1)`

If G is the point whose position vector is `barg`, then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE, CF internally in the ratio 2 : 1.

Therefore, three medians are concurrent.

L.H.S = `[(bara + barb,  barb + barc,  barc + bara)]`

= `(bara + barb) . [(barb + barc) xx (barc + bara)]`

= `(bara + barb) . [barb xx barc + barb xx bara + barc xx barc + barc xx bara]`

= `(bara + barb) . [barb xx barc + barb xx bara + barc xx bara]   ...[∵ barc xx barc = bar0]`

= `bara . [(barb xx barc) + (barb xx bara) + (barc xx bara)] + barb . [(barb xx barc) + (barb xx bara) + (barc xx bara)]`

= `bara . (barb xx barc) + bara . (barb xx bara) + bara . (barc xx bara) + barb . (barb xx barc) + barb(barb xx bara) + barb(barc xx bara)`

= `[bara  barb  barc] + [bara  barb  bara] + [bara  barc  bara] + [barb  barb  barc] + [barb  barb  bara] + [barb  barc  bara]`

= `[bara  barb  barc] + 0 + 0 + 0 + 0 + [bara  barb  barc]`

= `2[bara  barb  barc]`

= R.H.S

Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B.

Then ∠APB is an angle subtended on a semicircle.

Let `bar"AC" = <sup>bar"CB"</sup> = bar"a"` and `bar"CP" = bar"r"`

Then `|bar"a"| = |bar"r"|`       ....(1)

`bar"AP" = bar"AC" + bar"CP"`

= `bar"a" + bar"r"`

= `bar"r" + bar"a"`

`bar"BP" = bar"BC" + bar"CP"`

= `- bar"CB" + bar"CP"`

= `- bar"a" + bar"r"`

∴ `bar"AP".bar"BP" = (bar"r" + bar"a").(bar"r" - bar"a")`

= `bar"r".bar"r" - bar"r".bar"a" + bar"a".bar"r" - bar"a".bar"a"`

= `|bar"r"|^2 - |bar"a"|^2`

= 0    ....`(∵ bar"r".bar"a" = bar"a".bar"r")`

∴ `bar"AP" ⊥ bar"BP"`

∴ ∠APB is a right angle.

Hence, the angle subtended on a semicircle is the right angle.

Consider the circle with the centre at O and AB is the diameter.

Let `bar(OA) = bar a, bar(OB) = bar b, bar(OC) = bar c`

∴ `|bar a| =|bar b| = |bar c| = r`    ...(1)

and `bar a = -bar b`    ...(2)

Consider:

`bar (AC) * bar (BC) = (bar c - bar a) * (bar c - bar b)`

= `(bar c - bar a) * (bar c + bar a)`    ...[From (2)]

= `|bar c|^2 - |bar a|^2`

= r² − r²    ...[From (1)]

= 0

∴ `bar(AC) * bar(BC) = 0`

∴ `bar(AC)` is perpendicular to `bar(BC)`

∴ ∠ACB = 90°

∴ Angle subtended on semi-circle is a right angle.

1. Component Method: Resultant R = A + B is found as R_x = A_x + B_x​, R_y = A_y + B_y, R_z = A_z + B_z​, giving R = R_x\[\hat i\] + R_y\[\hat j\] + R_z\[\hat k\].

2. Laws of Addition: Triangle law (head-to-tail), Parallelogram law (tail-to-tail, diagonal = resultant), and Polygon law (for multiple vectors, closing side = resultant).

3. Magnitude (Addition): When A and B are at angle θ, R = \[\sqrt{A^2+B^2+2AB\cos\theta}\].

4. Magnitude (Subtraction): Change the sign to minus — ∣R∣ = ​\[\sqrt{A^2+B^2-2AB\cos\theta}\].

5. Direction of Resultant: tan⁡α = \[\frac{B\sin\theta}{A+B\cos\theta}\] for addition; tan⁡β = \[\frac{B\sin\theta}{A-B\cos\theta}\] for subtraction.

Special Cases

• Perpendicular → \[\overline{\mathrm{a}}\cdot\overline{\mathrm{b}}=0\]

• Parallel → \[\mathbf{\overline{a}}\cdot\mathbf{\overline{b}}=\mathbf{ab}\]

Projection

• Scalar= \[\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\]

• Vector \[=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^{2}}\cdot\mathbf{b}\]