Class 12 Mathematics Chapter 10 Vector Algebra

Key Concept: Vector Addition Equivalence

a) Both laws result in the same vector sum
[Solution Description] Both the triangle and parallelogram laws deal with vector addition and give the same result for the sum of two vectors. When two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ are added using either method, the resultant is the same vector $$\mathbf{c}$$. The triangle law visualizes this by placing the tail of one vector at the head of another, while the parallelogram law uses both vectors as adjacent sides of a parallelogram.

Key Concept: Coinitial Vectors

b) They originate from the same point
[Solution Description] Coinitial vectors share the same initial point but may have different directions and magnitudes.

Key Concept: Zero Vector

b) It has zero magnitude

[Solution Description] A zero vector is a vector that has both its magnitude and all its components equal to zero. It does not have a specific direction because its magnitude is zero. Hence, the statement that a zero vector has zero magnitude is correct.

Key Concept: Vector Magnitude

c) 13

[Solution Description]

The magnitude of a vector $$\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$$ is given by $$|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$$. Here, $$x = 4$$, $$y = -3$$, and $$z = 12$$.

Calculating the magnitude:

$$|\mathbf{r}| = \sqrt{(4)^2 + (-3)^2 + (12)^2}$$

=$$\sqrt{16 + 9 + 144}$$

= $$\sqrt{169}$$

= 13.

Thus, the magnitude of the vector is 13.

Key Concept: Simple Conversion

a) $$\left( \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$
[Solution Description]
Calculate the magnitude $$r$$ using:
$$
r = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt{2}.
$$
Then, the direction cosines are:
\begin{align*}
l &= \frac{6}{10\sqrt{2}} = \frac{3}{5\sqrt{2}}, \\
m &= \frac{8}{10\sqrt{2}} = \frac{4}{5\sqrt{2}}, \\
n &= \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}.
\end{align*}
Hence, the direction cosines are $$\left( \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$.

Key Concept: Properties of Scalar Product

b) It is associative

[Solution Description] The scalar product (dot product) has specific properties including being commutative, distributive over vector addition, but it does not have an associative property with respect to multiplication like numbers do. Hence, the statement that it is associative is incorrect.

Key Concept:Collinearity Condition, Vector Transformation

b) 1
[Solution Description]
Vectors $$\mathbf{x}$$ and $$\mathbf{y}$$ are said to be identical if they have the same magnitude and direction. This occurs when each component of $$\mathbf{x}$$ matches with the corresponding component of $$\mathbf{y}$$.

Thus, for components:
$$\lambda(5\hat{i}) = 5\hat{i}\quad\implies\quad\lambda = 1$$

Simultaneously,
$$\lambda(\hat{j}) =\hat{j}\quad\implies\quad\lambda = 1$$

Hence, the scalar $$\lambda$$ that makes vectors identical is 1.

Key Concept: Vector Components

a) $$2, 3, -1$$
[Solution Description] The scalar components of a vector in the form $$\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$$ are x, y, and z. Given $$\mathbf{r} = 2\mathbf{i} + 3\mathbf{j} -\mathbf{k}$$, the scalar components are $$x = 2$$, $$y = 3$$, and $$z = -1$$.

Key Concept: Conceptual Differentiation

c) Force has both magnitude and direction
[Solution Description] Force has both magnitude and direction, while time has only magnitude.

Key Concept: Coinitial Vectors

c) (1, 2, 3) and (1, 2, 3)
[Solution Description] Coinitial vectors share the same initial point. In option (c), both vectors are identical and thus can be considered to originate from the same starting point, making them coinitial by definition.

Key Concept: Scalar and Vector Distinction, Perpendicular Vectors

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Consider two non-zero vectors $$\vec{a}$$ and $$\vec{b}$$. The assertion claims that these vectors’ cross product will be zero if they are collinear. If $$\vec{a}$$ and $$\vec{b}$$ are collinear, they can be expressed as $$\vec{a} = k\vec{b}$$, where $$k$$ is a scalar. Therefore, the cross product $$\vec{a} \times \vec{b}$$ simplifies to:

$$\vec{a} \times \vec{b} = (k\vec{b}) \times \vec{b} = k(\vec{b} \times \vec{b}) = k \cdot 0 = 0$$

Thus, the assertion is true.

Now, consider the reason stating that when two vectors are perpendicular, their dot product is zero. This is indeed correct because for vectors $$\vec{c}$$ and $$\vec{d}$$, if they are perpendicular:

$$\vec{c} \cdot \vec{d} = |\vec{c}| |\vec{d}| \cos(90^\circ) = 0$$

However, the reason does not explain why the cross product being zero implies the vectors’ collinearity but rather states a different condition about perpendicularity. Hence, while both statements are true individually, the reason is not explaining the assertion.

Key Concept: Vector addition

a) $$\left( \frac{13}{5}, \frac{-8}{5}, \frac{33}{5} \right)$$

[Solution Description]

Using the section formula,

$$\mathbf{r} = \frac{m \mathbf{b} + n \mathbf{a}}{m+n}$$

where $$m = 3, n = 2, \mathbf{a} = (-1, 2, 3), \mathbf{b} = (5, -4, 9)$$.

Calculating each component:

$$x = \frac{3 \times 5 + 2 \times (-1)}{3 + 2} = \frac{15 – 2}{5} = \frac{13}{5}$$

$$y = \frac{3 \times (-4) + 2 \times 2}{3 + 2} = \frac{-12 + 4}{5} = \frac{-8}{5}$$

$$z = \frac{3 \times 9 + 2 \times 3}{3 + 2} = \frac{27 + 6}{5} = \frac{33}{5}$$

Therefore, the position vector of R is $$\left(\frac{13}{5}, \frac{-8}{5}, \frac{33}{5}\right)$$.

Key Concept: Zero Vector Concept

c) $$\vec{0}$$

[Solution Description] A zero vector is a vector with zero magnitude and no specific direction. It is usually denoted as the vector $$\vec{0}$$.

Key Concept: Vector Notation

c) $$\vec{v}$$
[Solution Description] Vectors are typically denoted with an arrow on top or in bold notation. Among the choices, $$\vec{v}$$ indicates a vector.

Key Concept: Opposite Vector

a) -a
[Solution Description]
A vector that has the same magnitude as vector $$\vec{a}$$ but points in the opposite direction is represented as $$-\vec{a}$$.

Key Concept: Basic Properties

b) A zero vector
[Solution Description] The equation $$\vec{c} + (-\vec{c}) = \vec{0}$$ represents that $$\vec{0}$$ is the zero vector, having no magnitude or direction.

Key Concept:Basic Definition

c) The product of their magnitudes times the cosine of the angle between them
[Solution Description] The scalar product, also known as the dot product, of two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is defined mathematically by the formula:
$$\mathbf{a}
\cdot
\mathbf{b} = |
\mathbf{a}||
\mathbf{b}|
cos
\theta
$$
where $$|\mathbf{a}|$$ and $$|\mathbf{b}|$$ are the magnitudes of vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ respectively, and $$\theta$$ is the angle between them.

Key Concept: Zero Vector in Vector Addition

c) $$\mathbf{a}$$
[Solution Description] In vector addition, adding a zero vector $$\mathbf{0}$$ to any vector $$\mathbf{a}$$ results in the same vector $$\mathbf{a}$$. This is because the zero vector is the additive identity.

Key Concept: Basic Projection

d) $$\mathbf{a}\cdot\hat{p}$$
[Solution Description] The projection of a vector $$\mathbf{a}$$ on a line with a unit vector $$\hat{p}$$ is given by the dot product of $$\mathbf{a}$$ and $$\hat{p}$$, which is $$\mathbf{a}\cdot\hat{p}$$.

Key Concept: Basic Definition

c) A directed line segment from point A to B
[Solution Description] A vector is defined as a quantity that has both magnitude and direction, such as displacement represented by a directed line segment.

Key Concept:Vector Addition, Real-world Application

a) 500 km
[Solution Description]
Consider eastward movement as positive x-direction and northward as positive y-direction. The displacement vector can be represented by:
$$\mathbf{R} = 300\mathbf{i} + 400\mathbf{j}$$

Magnitude $$|\mathbf{R}|$$:$$
|\mathbf{R}| =\sqrt{300^2 + 400^2} =\sqrt{90000 + 160000} =\sqrt{250000} = 500\text{ km}$$

Key Concept: Unit Vector Calculation

b) $$\left(0.23, -0.78, 0.59\right)$$
[Solution Description] The unit vector $$\hat{a}$$ is given by $$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$. The magnitude $$|\vec{a}|$$ is calculated as:
$$
|\vec{a}| = \sqrt{7^2 + (-24)^2 + 18^2} = \sqrt{49 + 576 + 324} = \sqrt{949}.
$$
Hence, the unit vector $$\hat{a}$$ is:
$$
\hat{a} = \left(\frac{7}{\sqrt{949}}, \frac{-24}{\sqrt{949}}, \frac{18}{\sqrt{949}}\right).
$$
Approximating gives $$\left(0.23, -0.78, 0.59\right)$$.

Key Concept: Collinear Vectors

b) Are parallel to the same line

[Solution Description] Collinear vectors are those that are parallel to the same line, regardless of their magnitudes and directions.

Key Concept: Unit Vector

b) j
[Solution Description] A unit vector along any axis is a vector of magnitude 1 in the direction of that axis. Along the y-axis, the unit vector is denoted by j.

Key Concept: Component Equality

b) $$n = -3$$
[Solution Description]
Equating the second components of the vectors, we find n:

n = -3

Thus, n = -3.

Key Concept: Vector Joining Two Points, Magnitude

a) No such $$\alpha$$ exists
[Solution Description] Calculate magnitudes:
$$
|\mathbf{u}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}
$$

$$
|\mathbf{v}| = \sqrt{4^2 + 5^2 + \alpha^2} = \sqrt{16 + 25 + \alpha^2} = \sqrt{41 + \alpha^2}
$$

Equating the magnitudes:
$$
\sqrt{14} = \sqrt{41 + \alpha^2}
$$

Squaring both sides:
$$
14 = 41 + \alpha^2
$$

Solving for $$\alpha^2$$ gives:
$$
\alpha^2 = 14 – 41 = -27 \quad (\text{This results in no real solution})
$$

Hence there is no real value for $$\alpha$$ such that the magnitudes are equal.

Key Concept: Basic Vector Concept

c) It has a magnitude of 1

[Solution Description] A unit vector is a vector with a magnitude of 1. It is used to indicate direction.

Key Concept:Advanced Magnitude Problems, Unit vector derivation

b) 1
[Solution Description] A unit vector has a magnitude of 1. Compute the magnitude of $$\mathbf{a}$$ first:

$$|\mathbf{a}| =\sqrt{(5)^2 + (12)^2} =\sqrt{25 + 144} =\sqrt{169} = 13$$

Now, find the unit vector:
$$\hat{a} =\frac{\mathbf{a}}{|\mathbf{a}|} =\frac{5}{13}\mathbf{i} +\frac{12}{13}\mathbf{j}$$

The magnitude of $$\hat{a}$$ is 1 since it is a unit vector.

Key Concept: Advanced Projection, Scalar Product

a) $$(6.5, 11.36, 0)$$
[Solution Description]
The vector with direction cosines $$(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0)$$ can be expressed as $$\mathbf{D} = (a, b, c)$$ where $$a = \frac{1}{2}$$, $$b = \frac{\sqrt{3}}{2}$$, $$c = 0$$.
The projection of $$\mathbf{C}$$ onto $$\mathbf{D}$$ is calculated using the formula:
$$\text{Projection of } \mathbf{C} \text{ on } \mathbf{D} = \frac{\mathbf{C} \cdot \mathbf{D}}{\|\mathbf{D}\|^2} \times \mathbf{D}$$
First calculate $$\mathbf{C} \cdot \mathbf{D}$$:
$$\mathbf{C} \cdot \mathbf{D} = 6 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 10 \cdot 0 = 3 + 4\sqrt{3}$$
Next, calculate $$\|\mathbf{D}\|^2$$
$$\|\mathbf{D}\|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + 0 = \frac{1}{4} + \frac{3}{4} = 1$$
Therefore, the projection becomes:
$$= \frac{3 + 4\sqrt{3}}{1} \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right)$$

$$= \left(\frac{3 + 4\sqrt{3}}{2}, \frac{(3 + 4\sqrt{3})\sqrt{3}}{2}, 0 \right)$$
Simplifying yields:
Projection ≈ $$(6.5, 11.36, 0)$$.

Choices below provide rounded or simplified estimates.

Key Concept: Opposite Vectors

c) $$-\vec{a}$$
[Solution Description] A vector that has the same magnitude but opposite direction to another vector $$\vec{a}$$ is given by $$-\vec{a}$$.

Key Concept: Angle Calculation

c) $$60^\circ$$
[Solution Description]
To find the angle $$\theta$$ between $$\mathbf{a}$$ and $$\mathbf{b}$$, we use the formula for the scalar product:
$$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta.$$
Plug in the values:
$$10 = 5\times 4\times \cos\theta.$$
Simplify this to get:
$$cos\theta =\frac{10}{20} = 0.5.$$
Thus,$$\theta =arccos(0.5) = 60^\circ.$$

Key Concept: Vector Definition

c) A quantity with both magnitude and direction
[Solution Description] A vector is defined as a quantity that has both magnitude and direction.

Key Concept:  Internal Division

b) $$\frac{m\mathbf{P} + n\mathbf{Q}}{m+n}$$
[Solution Description]
If a point divides a line segment internally in the ratio $$m:n$$, the position vector of such a point can be determined using the section formula: $$\frac{m\mathbf{P} + n\mathbf{Q}}{m+n}$$, where $$\mathbf{P}$$ and $$\mathbf{Q}$$ are the position vectors of the endpoints.

Key Concept: Basic Definition

c) $$a\cdot b = |a||b|\cos(\theta)$$
[Solution Description] The scalar product of two vectors $$a$$ and $$b$$ is defined by the formula $$a\cdot b = |a||b|\cos(\theta)$$, where $$\theta$$ is the angle between the vectors.

Key Concept: Vector Direction, Advanced Reasoning

c) $$77.87^\circ$$
[Solution Description]
The direction cosine with respect to the z-axis can be found using:

$$
\cos(\theta_z) = \frac{k}{|\mathbf{v}|},
$$

where $$k=1$$ and $$\mathbf{v}=\langle 3, -4, 1 \rangle$$. First, compute the magnitude of $$\mathbf{v}$$:
$$
|\mathbf{v}| = \sqrt{3^2 + (-4)^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}.
$$

Now find the cosine:
$$
\cos(\theta_z) = \frac{1}{\sqrt{26}}.
$$

Thus,
$$
\theta_z = \cos^{-1}\left(\frac{1}{\sqrt{26}}\right).
$$

Calculate this to find the angle approximately equals $$77.87^\circ$$.

Key Concept: Direction Ratios

a) Yes
[Solution Description] Compute direction ratios as $$\frac{-6}{3} = \frac{-12}{6} = \frac{-18}{9} = -2$$. The equal ratios confirm that the vectors are collinear.

Key Concept: Basic Properties

b) Distributivity over addition
[Solution Description] The distributivity of vector product over addition states that $$A\times (B + C) = A\times B + A\times C$$.

Key Concept: Right-Hand Rule

b) Negative z-axis
[Solution Description] Using the right-hand rule, point your fingers in the direction of $$\mathbf{A}$$ (positive x-axis) and curl them towards $$\mathbf{B}$$ (negative y-axis). Your thumb will point in the negative z-axis direction.

Key Concept: Unit Vector Components

b) $$\hat{i}, \hat{j}, \hat{k}$$
[Solution Description] Unit vectors along the x, y, and z axes are denoted by $$\hat{i}$$, $$\hat{j}$$, and $$\hat{k}$$ respectively. These symbols uniquely identify the standard coordinate axis directions.

Key Concept: Cross Product Properties

c) $$\mathbf{a}\times\mathbf{b} =\mathbf{b}\times\mathbf{a}$$
[Solution Description] All statements are true except for the one that suggests commutativity. The cross product is non-commutative, meaning $$\mathbf{a}\times\mathbf{b}\neq\mathbf{b}\times\mathbf{a}$$.

Key Concept: Perpendicular Vectors

a) $$\mathbf{a}\cdot\mathbf{b} = 0$$
[Solution Description] Two nonzero vectors are perpendicular when their scalar product is zero. This condition implies that the cosine of the angle between them is zero, which occurs at an angle of $$\pi/2$$ radians or 90 degrees.

Key Concept: Scenario-Based Question

b) Coinitial
[Solution Description] Since both $$\vec{A}$$ and $$\vec{B}$$ start from the origin, they qualify as coinitial vectors, which means they share the same initial point.

Key Concept: Basic Definition

c) $$\mathbf{a}\times\mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\hat{n}$$
[Solution Description] The vector product of two nonzero vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is given by the formula $$\mathbf{a}\times\mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\hat{n}$$, where $$\theta$$ is the angle between the vectors and $$\hat{n}$$ is a unit vector perpendicular to both.