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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Revision: Applications of Vector Algebra Mathematics HSC Science Class 12 Tamil Nadu Board of Secondary Education

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Definitions [8]

Definition: Magnitude of a Vector

The magnitude of vector \[\vec{AB}\] is the length of the directed line segment AB. It is written as \[|\vec{AB}|\], \[|\vec{a}|\], or simply a. The magnitude of a vector is never negative because it represents length.

Position Vector

In three-dimensional geometry, 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 written as \[\vec{OP}\]. If point P(x, y, z) is given, then the magnitude of its position vector is:

\[|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}\]
Definition: Scalar Quantity

A scalar quantity is a physical quantity that has magnitude only.

Definition: Vector Quantity

A vector quantity is a physical quantity that has magnitude as well as direction.

Definition: Vector

A vector is a quantity that has magnitude as well as direction. Geometrically, a vector is represented by a directed line segment such as  \[\vec{AB}\], where A is the initial point and B is the terminal point.

Definition: Scalar Product

The scalar product or dot product of two nonzero vectors \[\vec P\] and \[\vec Q\] is defined as the product of the magnitudes of the two vectors and the cosine of the angle θ between the two vectors.

Definition: Scalar Triple Product

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].

Definition: Vector Triple Product

For vectors \[\overline{a}\], \[\overline{b}\] and \[\overline{c}\] in the space, we define the vector triple product as

\[\overset{-}{\operatorname*{\operatorname*{a}}}\times\left(\overset{-}{\operatorname*{\operatorname*{b}}}\times\overset{-}{\operatorname*{\operatorname*{c}}}\right)=\left(\overset{-}{\operatorname*{\operatorname*{a}}}\cdot\overset{-}{\operatorname*{\operatorname*{c}}}\right)\overline{b}-\left(\overset{-}{\operatorname*{\operatorname*{a}}}\cdot\overline{b}\right)\overline{c}\]

Formulae [2]

Formula: Scalar (Dot) Product

\[\vec{A}\cdot\vec{B}=|\vec{A}||\vec{B}|\cos\theta=AB\cos\theta\]

Formula: Volume

Parallelepiped: Volume = [a b c]

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

Theorems and Laws [2]

Using properties of scalar triple product, prove that `[(bara + barb,  barb + barc,  barc + bara)] = 2[(bara, barb, barc)]`.

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

Prove by vector method, that the angle subtended on semicircle is a right angle.

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" = bar"CB" = 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.

Key Points

Key Points: Basic Concepts of Vector Algebra
  • Scalars have only magnitude.

  • Vectors have magnitude and direction.

  • Vectors are represented by directed line segments.

  • \[\vec{AB}\] represents a vector from A to B.

  • Magnitude of a vector is its length and is always non-negative.

  • \[\vec{OP}\] is the position vector of point \[P(x, y, z)\].

  • \[|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}\].

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