Revision: Vectors and Three-dimensional Geometry >> Vectors Maths Commerce (English Medium) Class 12 CBSE

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

The angles made by a vector with the positive directions of the X-axis, Y-axis and Z-axis are called direction angles of the vector, denoted by α, β, and γ.

If α, β and γ are the direction angles of a vector, then the cosines of these angles, i.e.

l = cos⁡α, m = cos⁡β, n = cos⁡γ 

are called the direction cosines of the vector.

If point is (x,y,z)and distance r: \[\cos\alpha=\frac{x}{r},\quad\cos\beta=\frac{y}{r},\quad\cos\gamma=\frac{z}{r}\]

If l, m, n are direction cosines of a line and if a, b, c are real numbers such that \[\frac{\mathrm{a}}{l}=\frac{\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{c}}{\mathrm{n}}=\lambda,\] then a, b, c are called direction ratios of that line.

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].

The scalar product or inner product of two non-zero vectors written as like \[\mid a\mid\mid b\mid\cos\theta\]\[\vec{a}\], \[\vec{b}\] is defined to be the scalar \[\left|\vec{a}\right|\left|\vec{b}\right|\cos\theta\] = \[ab\cos\theta\]

where a \[=|\vec{a}|\], b = \[=|\vec{b}|\] and θ = (0 ≤ θ ≤ π) is the angle between\[\vec{a}\] and \[\vec{b}\].

Let \[\hat{i}\],\[\hat{j}\], \[\hat{k}\] be unit vectors in the positive direction of the three mutually perpendicular coordinate axes, x-axis,  y-axis and z-axis, respectively. Then, these vectors are said to form an orthonormal triad of vectors. 

Dot Products: 

• \[\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1\]

• \[\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0\]

If a vector \[\overrightarrow{AB}\] is denoted by \[\overrightarrow{a}\], then \[\mid\overrightarrow{a}\mid\] denotes the positive length of the vector a, also called the magnitude or norm or modulus of the vector.

Thus \[\left|\vec{a}\right|\] = a, if a is the positive length of \[\overrightarrow{a}\].

\[\mid\overset{\rightarrow}{\operatorname*{\mathbf{AB}}}\mid=\mid\overset{\rightarrow}{\operatorname*{a}}\mid=a\]

A quantity which has both magnitude and direction is called a vector quantity, provided that two such quantities can be combined by vector addition.

A vector quantity can be represented in the plane by a directed line segment, whose length is proportional to the magnitude of the vector and whose direction is the direction of the vector.

When the direction of rotation is anticlockwise, then the rotation will move the screw upwards. It is called a right-handed orientation or a right-handed screw rule. 

Two vectors \[\vec{a}\] and \[\vec{b}\] are parallel if one is a scalar multiple of the other.

\[\vec{a}=\lambda\vec{b}\] (λ is a scalar)

The cosines of the angles made by a vector with the positive directions of the coordinate axes are called the direction cosines of the vector.

If a vector \[\vec{a}\] makes angles α,β,γ with the positive x, y and z axes respectively, then:

l = cosα, m = cosβ, n = cosγ

are called the direction cosines of the vector.

In Cartesian Form:

\[l=\frac{x}{r},\quad m=\frac{y}{r},\quad n=\frac{z}{r}\]

For any three given vectors, the scalar product of one of the vectors and the cross product of the remaining two, is called a scalar triple product

Thus, \[\vec{a},\vec{b},\vec{c}\] are three vectors, then \[(\vec{a}\times\vec{b})\cdot\vec{c}\]is called the scalar triple product and is denoted by \[[\vec{a}\vec{b}\vec{c}]\mathrm{~or~}[a,b,c]\]

Any three numbers proportional to direction cosines are called direction ratios.

• Denoted by a, b, c

• A line has infinitely many direction ratios.

\[l=\frac{a}{\sqrt{a^2+b^2+c^2}}\], \[m=\frac{b}{\sqrt{a^2+b^2+c^2}}\], \[n=\frac{c}{\sqrt{a^2+b^2+c^2}}\]

Let\[\vec{a}\] and \[\vec{b}\]be two non-zero, non-parallel vectors, and let θ be the angle between them such that (0 < θ < π).

\[\vec{a}\times\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\sin\theta\left.\hat{n}\right.\]

or

\[\vec{a}\times\vec{b}=ab\sin\theta\mathrm{~}\hat{n}\]

where \[\hat{n}\] is a unit vector perpendicular to both \[\vec{a}\] and\[\vec{b}\] such that\[\vec{a}\], \[\vec{b}\], \[\hat{n}\] form a righthanded triad of vectors.

 The square of a vector a, i.e., \[\vec{a^2}\] is a scalar which denotes the square of the length of a and is equal to the square of its modulus.

\[\vec{a^2}\] = \[|\vec{a}|^2\] 

A directed line segment is a line segment with an arrowhead showing direction. Its two endpoints are distinguishable as the initial point and the terminal point

The vector is denoted by \[\overrightarrow{AB}\]  

The vector drawn from the origin O(0,0,0)to a point P(x,y,z) is called the position vector of the point P.

It is denoted by: \[\vec{OP}=x\hat{i}+y\hat{j}+z\hat{k}\]

Magnitude of Position Vector: \[|\vec{OP}|=\sqrt{x^2+y^2+z^2}\]

When quantities can be represented by a certain number of units with no association with direction in space, they are called scalar quantities and numbers that represent them are called scalars. 

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

Parallelepiped: Volume = [a b c]

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

\[\lambda(a\hat{i}+b\hat{j})=\lambda a\hat{i}+\lambda b\hat{j}\]

\[\vec{OR}=\frac{m\vec{q}+n\vec{p}}{m+n}\]

\[\vec{OR}=\frac{m\vec{q}-n\vec{p}}{m-n}\]

Projection vector of \[\vec{a} on \vec{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}, \vec{b} \neq \vec{0}\]

Projection vector of \[\vec{b} on \vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}, \vec{a} \neq \vec{0}\]

cosθ \[= \frac{ \vec{a} \cdot \vec{b} }{ | \vec{a} | \, | \vec{b} | } = \frac{ \text{scalar product of the two vectors} }{ \text{product of their moduli} }\]

\[\vec{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}\]

\[|\vec{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

\[\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}\]

In 2D:

If \[\vec{a}=a_1\hat{i}+a_2\hat{j},\quad\vec{b}=b_1\hat{i}+b_2\hat{j}\]

\[\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2\]

Angle Between Two Vectors (2D):

\[\cos\theta=\frac{a_1b_1+a_2b_2}{\left|\vec{a}\right|\left|\vec{b}\right|}\]

In 3D

If  \[\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\quad\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\]

\[\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3\]

1. Vector area of a triangle:

\[\text{Vector area of }\triangle ABC=\frac{1}{2}(\vec{AB}\times\vec{AC})\]

2. Collinearity condition

\[\vec{a}\times\vec{b}+\vec{b}\times\vec{c}+\vec{c}\times\vec{a}=\vec{0}\]

3. Area Parallelogram

\[\text{Area of parallelogram}=|\vec{a}\times\vec{b}|\]

If \[\mathrm{M}({\overline{m}})\] is the mid-point of the line segment joining the points \[\mathrm{A}({\overline{a}})\] and \[\mathrm{B}({\overline{b}})\] then \[\overline{m}=\frac{\left(\overline{a}+\overline{b}\right)}{2}\]

Scalar projection = \[\frac{\text{scalar product}}{\text{Modulus of vector}}\] 

\[\text{Scalar Projection of }\overline{b}\mathrm{~on~}\overline{a}=\frac{\overline{a}\cdot\overline{b}}{|\overline{a}|}\]

\[\text{Vector Projection of }\overline{b}\mathrm{~on~}\overline{a}=\left(\overline{a}\cdot\overline{b}\right)\frac{\overline{a}}{\left|\overline{a}\right|^{2}}\]

\[(a\hat{i}+b\hat{j})+(x\hat{i}+y\hat{j})=(a+x)\hat{i}+(b+y)\hat{j}\]

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.

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`. 

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.

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)`

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.

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

Statement: 

If\[\vec{a},\vec{b},\vec{c},\] are three non-zero vectors and \[\vec{a}\times\vec{c}=\vec{b}\times\vec{c}\], then either \[\vec{a}=\vec{b}\mathrm{~or~}(\vec{a}-\vec{b})\] and \[\vec{c}\]are parallel vectors. 

\[\vec{a}\times\vec{c}=\vec{b}\times\vec{c}\Rightarrow\vec{a}=\vec{b}\mathrm{~or~}(\vec{a}-\vec{b})\parallel\vec{c}\]

The cross product has no cancellation law

If three points O, A, and B are so chosen that  \[\overrightarrow{OA}\] and  \[\overrightarrow{AB}\] respectively represent \[\overrightarrow{a}\] and \[\overrightarrow{b}\], then \[\overrightarrow{OB}\] is defined as the sum of  \[\overrightarrow{a}\] and \[\overrightarrow{b}\] and is written as \[\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}\], where  \[\overrightarrow{c}\] stands for the vector \[\overrightarrow{OB}\]. \[\overrightarrow{c}\] or \[\overrightarrow{a}\] + \[\overrightarrow{b}\] is also called the resultant of  \[\overrightarrow{a}\] and \[\overrightarrow{b}\]. This is known as the Triangle law of vectors.

The result of adding two co-initial vectors is the vector represented by the diagonal of the parallelogram formed with the component vectors as adjacent sides. This is the Parallelogram Law of addition of vectors, which is thus a direct consequence of the triangle law.

Statement:

Two vectors in a plane are equal iff their x-components and y-components are equal.

If \[\vec{a}=a_1\hat{i}+a_2\hat{j},\quad\vec{b}=b_1\hat{i}+b_2\hat{j}\] Then \[\vec{a}=\vec{b}\] ⟺a_{1 ​}= b₁​ and a₂​ = b₂​

1. Conversion: From D.R → D.C: \[l=\frac{a}{\sqrt{a^2+b^2+c^2}},m=\frac{b}{\sqrt{a^2+b^2+c^2}},n=\frac{c}{\sqrt{a^2+b^2+c^2}}\]

2. Angle between two lines

If direction cosines: \[\cos\theta=l_1l_2+m_1m_2+n_1n_2\]

If direction ratios: \[\cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\]

3. If A(x₁, y₁, z₁), B(x₂, y₂, z₂):

\[\mathrm{D.Rs}=(x_2-x_1,y_2-y_1,z_2-z_1)\]

4. \[l^2+m^2+n^2=1\]

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.

If\[\vec{a}\] and \[\vec{b}\] are two non-zero vectors, then \[\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\]

Cases:

• Acute angle (0< θ < \[\frac{\pi}{2}\])
  cos⁡θ > 0  ⇒  \[\vec{a}\cdot\vec{b}>0\]
  

• Right angle (θ = \[\frac{\pi}{2}\]
  cos⁡θ = 0  ⇒  \[\vec{a}\cdot\vec{b}=0\]
  

• Obtuse angle (\[\frac{\pi}{2}\] < θ ≤ π)

  cos⁡θ < 0  ⇒  \[\vec{a}\cdot\vec{b}<0\]

• Position of dot & cross doesn’t matter
  \[(\vec{a}\times\vec{b})\cdot\vec{c}=\vec{a}\cdot(\vec{b}\times\vec{c})\]
  

• Cyclic order unchanged ⇒ STP unchanged
  \[[\vec{a}\operatorname{\vec{b}}\vec{c}]=[\vec{b}\operatorname{\vec{c}}\vec{a}]=[\vec{c}\operatorname{\vec{a}}\vec{b}]\]
  

• Interchanging two vectors changes the sign
  \[[\vec{a}\operatorname{\vec{b}}\vec{c}]=-\left[\vec{b}\operatorname{\vec{a}}\vec{c}\right]\]
  

• If any two vectors are equal
  \[[\vec{a}\operatorname{\vec{a}}\vec{b}]=0\]

• If any two vectors are parallel
  \[[\vec{a}\operatorname{\vec{b}}\operatorname{\vec{c}}]=0\]

If: \[\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\]

Then:

Addition: \[\vec{a}+\vec{b}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k}\]

Scalar Multiplication: \[\lambda\vec{a}=\lambda a_1\hat{i}+\lambda a_2\hat{j}+\lambda a_3\hat{k}\]

If \[\vec{a}\] and \[\vec{b}\] are two vectors,

\[\vec{a}-\vec{b}=\vec{a}+(-\vec{b})\]

Let\[\vec{a}\] be a vector and m a scalar. Then m\[\vec{a}\] is called the product of  \[\vec{a}\] by the scalar m.

Properties:

• The direction of m\[\overrightarrow{a}\] is the same as or parallel to that of \[\overrightarrow{a}\].

• The magnitude of m\[\overrightarrow{a}\] is given by

  \[|m\vec{a}|=|m||\vec{a}|\]

• The sense of m→a is:

  • same as \[\vec{a}\], if m is positive

  • opposite to \[\vec{a}\], if m is negative