Vector Algebra

Vector Algebra deals with quantities that have both magnitude and direction. It provides an algebraic method to solve problems in geometry and physics. The chapter starts with the definition of vectors and scalars, vector addition using the Triangle and Parallelogram laws, and the multiplication of a vector by a scalar. You will learn to use position vectors to define points in space and apply the section formula to find points that divide a line segment in a given ratio.

The main part of the chapter covers two types of vector multiplication. The dot product (or scalar product) gives a scalar value and is used to find the angle between two vectors or the projection of one vector onto another. The cross product (or vector product) yields a new vector perpendicular to the plane of the original two, and its magnitude helps calculate the area of a triangle or parallelogram. These products are the primary tools for solving geometric problems.

Finally, the chapter introduces the scalar triple product (STP) and the vector triple product (VTP). The STP is used to find the volume of a parallelepiped and to check if three vectors are coplanar. The chapter gets difficult when applying these products to complex 3D geometry problems, which requires strong visualization skills. It relies heavily on concepts from basic coordinate geometry and trigonometry.

Formulas for Vector Algebra

Magnitude of a vector: $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$ For a vector \vec{a} = x\hat{i} + y\hat{j} + z\hat{k}.
Section Formula for internal division: $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$ \vec{r} is the position vector of a point dividing the line joining \vec{a} and \vec{b} in the ratio m:n.
Dot Product (Scalar Product): $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$ \theta is the angle between vectors \vec{a} and \vec{b}.
Dot Product in component form: $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ For \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}.
Scalar Projection of one vector on another: $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ This gives the length of the projection of vector \vec{a} onto vector \vec{b}.
Cross Product (Vector Product): $\vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin\theta) \hat{n}$ \hat{n} is a unit vector perpendicular to the plane of \vec{a} and \vec{b}.
Cross Product in determinant form: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ A practical way to compute the cross product of vectors in component form.
Area of a Triangle: $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|$ Where \vec{a} and \vec{b} are two adjacent sides of the triangle.
Scalar Triple Product (STP): $[\vec{a} \ \vec{b} \ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$ Its magnitude is the volume of the parallelepiped with adjacent edges \vec{a}, \vec{b}, \vec{c}.
Condition for Coplanarity: $[\vec{a} \ \vec{b} \ \vec{c}] = 0$ Three non-zero vectors are coplanar if their scalar triple product is zero.
Vector Triple Product (VTP): $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ The resultant vector lies in the plane containing \vec{b} and \vec{c}.

One of these questions, worked through

JEE Advanced 2022

Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let \ $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ , \ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ , $b_{2}, b_{3} \in \mathbb{R}$ \ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ , $c_{1}, c_{2}, c_{3} \in \mathbb{R}$ be three vectors such that $b_{2}b_{3} > 0$ , $\vec{a} \cdot \vec{b} = 0$ and $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}.$ Then, which of the following is/are TRUE ?

Vector Algebra Let

\hat{i}, \hat{j}

and

\hat{k}

be the unit vectors along the three positive coordinate axes. Let

\vec{a} = 3\hat{i} + \hat{j} - \hat{k}

\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}

b_{2}, b_{3} \in \mathbb{R}

\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}

c_{1}, c_{2}, c_{3} \in \mathbb{R}

be three vectors such that

b_{2}b_{3} > 0

\vec{a} \cdot \vec{b} = 0

and

\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}.

Then, which of the following is/are TRUE ?

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ $\text{Given: } b_{2}b_{3} > 0,\quad \vec{a} \cdot \vec{b} = 0$ $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}$ $\text{To find: which of (1) - (4) is/are TRUE}$ $\text{Plan: } (1)\ \text{Matrix} \Rightarrow \vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $(2)\ \text{dot } \vec{b} \Rightarrow \vec{b}\cdot\vec{c}=0\,;\quad (3)\ \text{dot } \vec{c} \Rightarrow \vec{a}\cdot\vec{c}=|\vec{c}|^2$ $(4)\ |\vec{b}|>\sqrt{10}\ \text{via } b_2b_3>0\,;\quad (5)\ |\vec{c}|\le|\vec{a}|=\sqrt{11}$ $\vec{a}\cdot\vec{b}=0 \Rightarrow \vec{a}\perp\vec{b}$ $[\vec{c}]_\times = \begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}$ $[\vec{c}]_\times\,\vec{b} = \vec{c}\times\vec{b}$ $(3-c_1,\ 1-c_2,\ -1-c_3) = \vec{a}-\vec{c}$ $\vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $\vec{b}\cdot(\vec{c}\times\vec{b}) = \vec{b}\cdot(\vec{a}-\vec{c})$ $0 = \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{c}$ $\vec{b}\cdot\vec{a} = 0 \Rightarrow \vec{b}\cdot\vec{c} = 0$ $\vec{b}\cdot\vec{c}=0 \quad \checkmark\ \text{(option 2 is TRUE)}$ $\vec{c}\cdot(\vec{c}\times\vec{b}) = \vec{c}\cdot(\vec{a}-\vec{c})$ $0 = \vec{c}\cdot\vec{a} - \vec{c}\cdot\vec{c} = \vec{a}\cdot\vec{c} - |\vec{c}|^2$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2$ $\vec{c}=0 \Rightarrow \vec{c}\times\vec{b}=0=\vec{a}-\vec{c}=\vec{a}\ \text{(false)}$ $\vec{a}\cdot\vec{c}=|\vec{c}|^2>0 \Rightarrow \text{option (1) is FALSE}$ $\vec{a}\cdot\vec{b} = 3(1) + 1(b_2) + (-1)(b_3) = 0$ $3 + b_2 - b_3 = 0$ $b_3 = b_2 + 3$ $b_2 b_3>0 \Rightarrow b_2(b_2+3)>0$ $b_2<-3 \ \text{ or } \ b_2>0$ $|\vec{b}|^2 = 1+b_2^2+b_3^2 = 1+b_2^2+(b_2+3)^2$ $|\vec{b}|^2 = 2b_2^2+6b_2+10$ $|\vec{b}|^2-10 = 2b_2(b_2+3) = 2b_2b_3$ $|\vec{b}|^2-10 = 2b_2b_3 > 0 \Rightarrow |\vec{b}|^2 > 10$ $|\vec{b}| > \sqrt{10} \quad \checkmark\ \text{(option 3 is TRUE)}$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2 \le |\vec{a}|\,|\vec{c}|$ $|\vec{c}| \le |\vec{a}| = \sqrt{9+1+1} = \sqrt{11}$ $|\vec{c}| \le \sqrt{11} \quad \checkmark\ \text{(option 4 is TRUE)}$ $\vec{c}\ \text{lies inside the sphere of radius}\ |\vec{a}|=\sqrt{11}$ $\text{Answer: } (2),\ (3),\ (4)$ $\vec{b}\cdot\vec{c}=0\,;\quad |\vec{b}|>\sqrt{10}\,;\quad |\vec{c}|\le\sqrt{11}$

Open this question on its own page →

Past-paper questions on Vector Algebra

JEE Advanced 30 questions