Vector Algebra Ex 10.2 Class 12 NCERT — Concepts & Solutions

Core Algebraic Concepts in Exercise 10.2

Vector Components

The values x, y, and z in the expression a = xi + yj + zk are called scalar components, while xi, yj, and zk are called the vector components along the coordinate axes.

Magnitude of a Vector

The length or size of a vector, computed using the Pythagorean theorem in 3D: |a| = sqrt(x² + y² + z²).

Unit Vector

A vector whose magnitude is exactly 1 unit. A unit vector in the direction of vector a is written as a-hat = a / |a|.

Section Formula

A formula used to find the position vector of a point dividing the line segment joining two points in a given ratio, either internally or externally.

Understanding Vector Components and Direction Cosines

When representing a vector algebraically, we decompose it into perpendicular components along the coordinate axes. For any vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ denote the positive directions of the X, Y, and Z axes respectively. Here, $x, y, z$ represent the scalar projections (or scalar components) of the vector along these axes. These components act as coordinates of the terminal point when the vector starts from the origin.

Direction cosines ($l, m, n$) represent the cosines of the angles $\alpha, \beta, \gamma$ that the vector makes with the positive X, Y, and Z axes. Mathematically, they are computed as $l = \frac{x}{|\vec{a}|}$, $m = \frac{y}{|\vec{a}|}$, and $n = \frac{z}{|\vec{a}|}$. A critical property to remember is that $l^2 + m^2 + n^2 = 1$. The scalar components $(x, y, z)$ are proportional to these direction cosines and are called direction ratios. Mastering this component form allows you to perform addition, subtraction, and scalar multiplication simply by adding, subtracting, or multiplying corresponding components.

How to Find the Unit Vector in the Direction of a Vector

1. Identify the Scalar Components — Locate the coefficients of i, j, and k in the vector expression vector a = xi + yj + z*k.
2. Calculate the Magnitude — Evaluate the magnitude |a| using the formula |a| = sqrt(x^2 + y^2 + z^2).
3. Divide the Vector by its Magnitude — Divide each scalar component of the vector by the calculated magnitude to obtain the unit vector elements.
4. Perform a Sanity Check — Ensure that the sum of squares of the new components equals 1, proving that the magnitude is indeed 1.

Board Exam Traps & Common Pitfalls

1. External Division Sign Error: In the section formula, students frequently write the internal division formula instead of the external one. For external division, always use a minus sign: $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$. Ensure you associate $m$ with the second point's vector $\vec{b}$ and $n$ with the first point's vector $\vec{a}$.
2. Magnitude Calculations: Don't forget that squaring a negative component makes it positive. For example, if $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$, then $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$ (not $\sqrt{4 - 9 + 16}$).
3. Collinearity Test: To prove two vectors $\vec{a}$ and $\vec{b}$ are collinear, show that $\vec{b} = \lambda \vec{a}$ for some scalar constant $\lambda$.

Practice Questions with Solutions

• Q: Find the unit vector in the direction of the sum of the vectors a = 2i + 2j - 5k and b = 2i + j + 3*k. A: Step 1: Find the sum vector c = a + b. c = (2 + 2)i + (2 + 1)j + (-5 + 3)k c = 4i + 3j - 2k Step 2: Calculate the magnitude of vector c. |c| = sqrt(4^2 + 3^2 + (-2)^2) = sqrt(16 + 9 + 4) = sqrt(29) Step 3: Write down the unit vector by dividing c by its magnitude. c-hat = (4/sqrt(29))i + (3/sqrt(29))j - (2/sqrt(29))k Final answer: The unit vector is (4/sqrt(29))i + (3/sqrt(29))j - (2/sqrt(29))k.
• Q: Find the vector joining the points P(2, 3, 0) and Q(-1, -2, -4) directed from P to Q. A: Step 1: Write down the position vectors of P and Q relative to the origin. OP = 2i + 3j + 0k OQ = -1i - 2j - 4k Step 2: Find the directed vector PQ by subtracting OP from OQ. PQ = OQ - OP PQ = (-1 - 2)i + (-2 - 3)j + (-4 - 0)k Step 3: Simplify the expression. PQ = -3i - 5j - 4k Final answer: The vector directed from P to Q is -3i - 5j - 4k.
• Q: Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are OP = i + 2j - k and OQ = -i + j + k respectively, in the ratio 2:1 externally. A: Step 1: Identify the position vectors and the ratio components. a = i + 2j - k, b = -i + j + k, m = 2, n = 1 Step 2: Apply the external division section formula: r = (mb - na) / (m - n). r = [2(-i + j + k) - 1(i + 2j - k)] / (2 - 1) Step 3: Expand and group the component terms. r = [-2i + 2j + 2k - i - 2j + k] / 1 r = -3i + 0j + 3k Final answer: The position vector of point R is -3i + 3k.
• Q: Show that the vectors a = 2i - 3j + 4k and b = -4i + 6j - 8k are collinear. A: Step 1: Write down the expression for vector b. b = -4i + 6j - 8k Step 2: Factor out a common scalar scalar multiplier from vector b. b = -2(2i - 3j + 4k) Step 3: Observe that the expression in the parenthesis is identical to vector a. b = -2 a Step 4: Conclude that since b is a scalar multiple of a (b = lambdaa where lambda = -2), the two vectors are parallel and collinear. Final answer: The vectors are collinear as b = -2*a.

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