This article contains , solution for Exercise 10.3 of Chapter 10 – Vector Algebra from Class 12 NCERT – Mathematics .Exercise 10.3 covers the basics of vectors like angle between vectors, magnitude of vectors, projection of vector, etc.

Exercise 10.3 of Chapter 10- Vector Algebra deals with the application of vectors in various operations and problems. such as Scalar (Dot) Product of Vectors and its properties, Projection of Vectors and finding the angle between vectors.

Important formulas for Chapter 10 – Vector Algebra – Exercise 10.3

These formulas will help you to solve the Exercise – 10.3 of Class 12 NCERT – Mathematics -Chapter 10 – Vector Algebra.

• Dot Product Formula:
  • [Tex]\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta[/Tex]
• Magnitude of a Vector:
  • [Tex]|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}[/Tex]
• Angle Between Two Vectors :
  • [Tex]\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}[/Tex]
  • [Tex]\theta = \cos^{-1} \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right)[/Tex]
• Projection of Vector :
  • [Tex]\vec{a}[/Tex] on [Tex]\vec{b}[/Tex]:[Tex]\text{Projection}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}[/Tex]
• Unit Vector:[Tex]\hat{u} = \frac{\vec{u}}{|\vec{u}|}[/Tex]
  
• Perpendicular Vectors: I
  • f [Tex]\vec{a} \cdot \vec{b} = 0[/Tex], then [Tex]\vec{a} \ and \ \vec{b}[/Tex] are perpendicular.
• Scalar Triple Product:
  • [Tex]\vec{a} \cdot (\vec{b} \times \vec{c}) = 0 [/Tex]This can be used to check for coplanarity of vectors.
• Vector Addition:
  • [Tex]\vec{a} + \vec{b} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j} + (a_z + b_z)\hat{k}[/Tex]
• Vector Subtraction:
  • [Tex]\vec{a} – \vec{b} = (a_x – b_x)\hat{i} + (a_y – b_y)\hat{j} + (a_z – b_z)\hat{k}[/Tex]
• Condition for Collinear Points:
  • The direction vectors between pairs of points should be proportional.

Class 12 NCERT Solutions- Mathematics – Chapter 10 – Vector Algebra Exercise 10.3

1. Find the angle between two vectors [Tex]\vec{a} [/Tex] and [Tex]\vec{b}[/Tex] with magnitudes √3 and 2, respectively having [Tex]\vec{a}.\vec{b}[/Tex] = √6.

Given:

[Tex]\| \vec{a} \| = \sqrt{3}[/Tex]

[Tex]\| \vec{b} \| = 2[/Tex]

[Tex]\vec{a} \cdot \vec{b} = \sqrt{6}[/Tex]

Substituting the given values into the formula:

[Tex]\cos \theta = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2}[/Tex]

To find the angle [Tex]\theta[/Tex], we take the inverse cosine of this value:

[Tex]\theta = \cos^{-1} \left( \frac{\sqrt{2}}{2} \right)[/Tex]

Using a calculator, [Tex]\cos^{-1} \left( \frac{\sqrt{2}}{2} \right)[/Tex] = π/4 or 45°.

So, the angle between the two vectors [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex] is 45°.

2. Find the angle between the vectors [Tex]\hat{i} – \hat{2j} + \hat{3k}[/Tex] and [Tex]\hat{3i} – \hat{2j} + \hat{k}[/Tex]

Given:

[Tex]\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k} [/Tex]

[Tex]\vec{b} = 3\hat{i} – 2\hat{j} + \hat{k}[/Tex]

To calculate the dot product, we multiply corresponding components and then add:

[Tex]\vec{a} \cdot \vec{b}[/Tex] = (1 · 3) + (-2 · -2) + (3 · 1) = 3 + 4 + 3 = 10

The magnitudes of each vector can be found using the formula for magnitude of a vector:

[Tex]\vec{a} = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}[/Tex]

[Tex]\vec{b} = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}[/Tex]

Now, substituting these values into the formula for the cosine of the angle between the vectors:

[Tex]\cos \theta = \frac{10}{\sqrt{14} \times \sqrt{14}} = \frac{10}{14} = \frac{5}{7}[/Tex]

To find the angle, we take the inverse cosine of this value:

? = cos^-1(5/7)

3. Find the projection of the vector [Tex]\hat{i} – \hat{j}[/Tex] on the vector [Tex]\hat{i} + \hat{j}[/Tex].

First, let’s find the dot product of [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex]:

[Tex]\vec{a} \cdot \vec{b}[/Tex] = (1 · 1) + (-1 · 1) = 1 – 1 = 0

Next, let’s find the magnitude of [Tex]\vec{b}[/Tex]:

[Tex]\| \vec{b}||[/Tex] = √{1² + 1²} = √{2}

Now, let’s find the projection:

[Tex]\text{Projection}_{\vec{b}}(\vec{a}) = \frac{0}{\sqrt{2}} \frac{\hat{i} + \hat{j}}{\sqrt{2}} = 0 \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = 0[/Tex]

So, the projection is the zero vector.

4. Find the projection of the vector [Tex]\hat{i} + \hat{3j} + \hat{7k}[/Tex] on the vector [Tex]\hat{7i} – \hat{j} + \hat{8k}[/Tex].

First, let’s find the dot product of [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex]:

[Tex]\vec{a} \cdot \vec{b}[/Tex] = (1 · 7) + (3 · -1) + (7 · 8) = 7 – 3 + 56 = 60

Next, let’s find the magnitude of [Tex]\vec{b}[/Tex]:

[Tex]\| \vec{b} \|[/Tex] = √{7² + (-1)² + 8²} = √{49 + 1 + 64} = √114.

Now, let’s find the projection:

[Tex]\text{Projection}_{\vec{b}}(\vec{a}) = \frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|} = \frac{60}{\sqrt{114} }[/Tex]

5. Show that each of the given three vectors is a unit vector:

[Tex](1/7)(\hat{2i} + \hat{3j} + \hat{6k})[/Tex], [Tex](1/7)(\hat{3i} – \hat{6j} + \hat{2k})[/Tex], [Tex](1/7)(\hat{6i} + \hat{2j} – \hat{3k})[/Tex]

Also, show that they are mutually perpendicular to each other.

For [Tex]\vec{v_1}[/Tex]:

[Tex]\| \vec{v_1} \| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2} = \sqrt{\frac{4+9+36}{49}} = \sqrt{\frac{49}{49}} = 1[/Tex]

Similarly, for [Tex]\vec{v_2}[/Tex]:

[Tex]\| \vec{v_2} \| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} = \sqrt{\frac{9+36+4}{49}} = \sqrt{\frac{49}{49}} = 1[/Tex]

And for [Tex]\vec{v_3}[/Tex]:

[Tex]\| \vec{v_3} \| = \sqrt{\left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2} = \sqrt{\frac{36+4+9}{49}} = \sqrt{\frac{49}{49}} = 1[/Tex]

So, each of the given vectors is indeed a unit vector.

Next, to show that they are mutually perpendicular, we can check if the dot products of each pair of vectors equal 0. If the dot product is 0, it means the vectors are orthogonal (perpendicular).

For [Tex]\vec{v_1}[/Tex] and [Tex]\vec{v_2}[/Tex]:

[Tex]\vec{v_1} \cdot \vec{v_2} = \left(\frac{2}{7}\right) \left(\frac{3}{7}\right) + \left(\frac{3}{7}\right) \left(-\frac{6}{7}\right) + \left(\frac{6}{7}\right) \left(\frac{2}{7}\right) = \frac{6}{49} – \frac{18}{49} + \frac{12}{49} = 0[/Tex]

For [Tex]\vec{v_1}[/Tex] and [Tex]\vec{v_3}[/Tex]:

[Tex]\vec{v_1} \cdot \vec{v_3} = \left(\frac{2}{7}\right) \left(\frac{6}{7}\right) + \left(\frac{3}{7}\right) \left(\frac{2}{7}\right) + \left(\frac{6}{7}\right) \left(-\frac{3}{7}\right) = \frac{12}{49} + \frac{6}{49} – \frac{18}{49} = 0[/Tex]

For [Tex]\vec{v_2}[/Tex] and [Tex]\vec{v_3}[/Tex]:

[Tex]\vec{v_2} \cdot \vec{v_3} = \left(\frac{3}{7}\right) \left(\frac{6}{7}\right) + \left(-\frac{6}{7}\right) \left(\frac{2}{7}\right) + \left(\frac{2}{7}\right) \left(-\frac{3}{7}\right) = \frac{18}{49} – \frac{12}{49} – \frac{6}{49} = 0[/Tex]

Since all dot products are zero, it confirms that the vectors are mutually perpendicular to each other.

6. Find |[Tex]\vec{a}[/Tex]| and |[Tex]\vec{b}[/Tex]|, if [Tex](\vec{a} + \vec{b}).(\vec{a} – \vec{b}) = 8[/Tex] and [Tex]|\vec{a}| = 8|\vec{b}|[/Tex].

[Tex](\vec{a} + \vec{b}).(\vec{a} – \vec{b}) = 8[/Tex]

[Tex]\vec{a}.\vec{a} – \vec{a}.\vec{b} + \vec{b}.\vec{a} – \vec{b}.\vec{b} = 8[/Tex]

[Tex]|\vec{a}|^2 – |\vec{b}|^2 = 8[/Tex]

[Tex]|8\vec{b}|^2 – |\vec{b}|^2 = 8[/Tex]

[Tex]64|\vec{b}|^2 – |\vec{b}|^2 = 8[/Tex]

[Tex]63|\vec{b}|^2 = 8[/Tex]

[Tex]|\vec{b}|^2[/Tex] = 8/63

[Tex]|\vec{b}|[/Tex] = √(8/63)

[Tex]|\vec{a}| = |\vec{b}|[/Tex] = 8 × √(8/63) = 16√2 / 3√7.

7. Evaluate the product [Tex](\vec{3a} – \vec{5b}).(\vec{2a} + \vec{7b})[/Tex].

[Tex](\vec{3a} – \vec{5b}).(\vec{2a} + \vec{7b}) = \vec{3a} \cdot \vec{2a} + \vec{3a} \cdot \vec{7b} – \vec{5b} \cdot \vec{2a} – \vec{5b} \cdot \vec{7b}[/Tex]

Now, apply the dot product:

[Tex](\vec{3a} – \vec{5b}).(\vec{2a} + \vec{7b}) = (3)(2)|\vec{a}|^2 + (3)(7)|\vec{a}||\vec{b}| – (5)(2)|\vec{a}||\vec{b}| – (5)(7)|\vec{b}|^2[/Tex]

[Tex]= 6|\vec{a}|^2 + 21|\vec{a}||\vec{b}| – 10|\vec{a}||\vec{b}| – 35|\vec{b}|^2[/Tex]

[Tex]= 6|\vec{a}|^2 + 11|\vec{a}||\vec{b}| – 35|\vec{b}|^2[/Tex]

8. Find the magnitude of two vectors [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex], having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

[Tex]\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)[/Tex]

where [Tex]\theta[/Tex] is the angle between the vectors.

Given that [Tex]|\vec{a}| = |\vec{b}|[/Tex], let’s denote this common magnitude as [Tex]|\vec{a}| = |\vec{b}| = x[/Tex].

Now, we know cos(60°) = 1/2, so we can rewrite the scalar product equation as:

1/2 = x · x · 1/2

1/2 = x²/2

1 = x²

x = 1

So, the magnitude of both vectors is 1.

9. Find |[Tex]\vec{x}[/Tex]|, if for a vector [Tex]\vec{a}[/Tex], [Tex](\vec{x} – \vec{a}).(\vec{x} + \vec{a}) = 12[/Tex]

[Tex](\vec{x} – \vec{a}).(\vec{x} + \vec{a}) = 12[/Tex]

[Tex]\vec{x}.\vec{x} + \vec{x}.\vec{a} – \vec{a}.\vec{x} – \vec{a}.\vec{a} = 12[/Tex]

[Tex]|\vec{x}|^2 – |\vec{a}|^2 = 12[/Tex]

[Tex]|\vec{x}|^2 – 1 = 12[/Tex]

[Tex]|\vec{x}|^2 = 13[/Tex]

[Tex]|\vec{x}|[/Tex] = √13

10. If [Tex]\vec{a} = \hat{2i} + \hat{2j} + \hat{3k}[/Tex], [Tex]\vec{b} = -\hat{i} + \hat{2j} + \hat{k}[/Tex] and [Tex]\vec{c} = \hat{3i} + \hat{j}[/Tex] are such that [Tex]\vec{a}[/Tex] + λ[Tex]\vec{b}[/Tex] is perpendicular to [Tex]\vec{c}[/Tex], then find the value of λ.

We’re given that a +λb is perpendicular to c. So, the dot product of should be zero:

[Tex](\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0[/Tex]

Expanding this expression:

[Tex](\vec{a} + \lambda \vec{b}) \cdot \vec{c} = (\vec{a} \cdot \vec{c}) + (\lambda \vec{b} \cdot \vec{c}) = 0[/Tex]

Substituting the given vectors:

[Tex](2\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + \hat{j}) + \lambda (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j}) = 0[/Tex]

(2⋅3) + (2⋅1) + (3⋅0) + λ ((−1⋅3) + (2⋅1) + (1⋅1)) = 0

6 + 2 + λ(−3 + 2 + 1) = 0

8 − λ = 0

λ = 8

So, the value of λ is 8.

11. Show that [Tex]|\vec{a}| \vec{b} + |\vec{b}| \vec{a}[/Tex] is perpendicular to [Tex]|\vec{a}| \vec{b} – |\vec{b}| \vec{a}[/Tex], for any two nonzero vectors [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex].

[Tex]\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 – |\vec{a}| |\vec{b}| |\vec{a}| |\vec{b}| + |\vec{b}| |\vec{a}| |\vec{a}| |\vec{b}| – |\vec{b}|^2 |\vec{a}|^2[/Tex]

[Tex]\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 – |\vec{a}| |\vec{b}| |\vec{a}| |\vec{b}| + |\vec{a}| |\vec{b}| |\vec{a}| |\vec{b}| – |\vec{b}|^2 |\vec{a}|^2[/Tex]

[Tex]\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 – |\vec{b}|^2 |\vec{a}|^2[/Tex]

[Tex]\vec{u} \cdot \vec{v} = 0[/Tex]

Therefore, [Tex]|\vec{a}| \vec{b} + |\vec{b}| \vec{a}[/Tex] is perpendicular to [Tex]|\vec{a}| \vec{b} – |\vec{b}| \vec{a}[/Tex].

12. If [Tex]\vec{a} . \vec{a} = 0[/Tex] and [Tex]\vec{a} . \vec{b} = 0[/Tex], then what can be concluded about the vector

If [Tex]\vec{a} \cdot \vec{a} = 0[/Tex] and [Tex]\vec{a} \cdot \vec{b} = 0[/Tex], it means that [Tex]\vec{a}[/Tex] is perpendicular to itself and also perpendicular to [Tex]\vec{b}[/Tex].

Since the dot product of [Tex]\vec{a}[/Tex] with itself is 0, it implies that the magnitude of [Tex]\vec{a}[/Tex] is 0. However, vectors with magnitude 0 are usually defined as the zero vector, denoted by [Tex]\vec{0}[/Tex].

Thus, the conclusion about [Tex]\vec{b}[/Tex] is that it can be any vector, as long as it is perpendicular to [Tex]\vec{a}[/Tex]. The only constraint is that [Tex]\vec{b}[/Tex] must be perpendicular to [Tex]\vec{a}[/Tex].

13. If [Tex]\vec{a}, \vec{b}, \vec{c}[/Tex] are unit vectors such that [Tex]\vec{a} + \vec{b} + \vec{c} = \vec{0}[/Tex], find the value of [Tex]\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}[/Tex].

Given: [Tex]\vec{a} + \vec{b} + \vec{c} = \vec{0}[/Tex] [Equation 1]and are unit vectors.

Multiplying equation 1 with vector a:

[Tex]\vec{a}[\vec{a} + \vec{b} + \vec{c}] = \vec{a}.\vec{0}[/Tex]

[Tex]\vec{a}.\vec{a} + \vec{a}.\vec{b} + \vec{a}.\vec{c} = \vec{a}.\vec{0}[/Tex]

[Tex]1+ \vec{a}.\vec{b} + \vec{a}.\vec{c} = 0[/Tex] …(a)

Multiplying equation 1 with vector b:

[Tex]\vec{b}.\vec{a} + 1 + \vec{b}.\vec{c} = 0[/Tex] …(b)

Multiplying equation 1 with vector c:

[Tex]\vec{c}.\vec{a} + \vec{c}.\vec{b} + 1 = 0[/Tex] …(c)

From (a), (b) and (c):

[Tex](1+ \vec{a}.\vec{b} + \vec{a}.\vec{c}) + (\vec{b}.\vec{a} + 1 + \vec{b}.\vec{c}) + (\vec{c}.\vec{a} + \vec{c}.\vec{b} + 1) = 0 + 0 + 0[/Tex]

[Tex](3 + \vec{a}.\vec{b} + \vec{c}.\vec{a}) + (\vec{a}.\vec{b} + \vec{b}.\vec{c}) + (\vec{c}.\vec{a} + \vec{b}.\vec{c}) = 0[/Tex]

[Tex]3 + 2(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}) = 0[/Tex]

[Tex]\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} = -3/2[/Tex]

14. If either vector [Tex]\vec{a} = \vec{0}[/Tex] or [Tex]\vec{b} = \vec{0}[/Tex], then [Tex]\vec{a}.\vec{b} = 0[/Tex]. But the converse need not be true. Justify your answer with an example

[Tex]\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta[/Tex], where [Tex]|\vec{a}|[/Tex] and [Tex]|\vec{b}|[/Tex] are the magnitudes of vectors [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex] respectively, and [Tex]\theta[/Tex] is the angle between them.

If either [Tex]\vec{a}[/Tex] or [Tex]\vec{b}[/Tex] is the zero vector, then the magnitude of that vector is 0. So, [Tex]|\vec{a}| = 0[/Tex] or [Tex]|\vec{b}| = 0[/Tex].

Thus, in such cases, [Tex]\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 0 \cdot |\vec{b}| \cos \theta = 0[/Tex], regardless of the value of [Tex]\vec{b}[/Tex].

However, the converse need not be true.

For example, let’s consider vectors [Tex]\vec{a}[/Tex] = (1, 0) and [Tex]\vec{b} = (0, 1)[/Tex].

Calculate the dot product: [Tex]\vec{a} \cdot \vec{b}[/Tex] = (1)(0) + (0)(1) = 0.

But neither a or b is the zero vector.

This example shows that a dot product of zero does not necessarily mean that one or both of the vectors are zero vectors.

15. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors [Tex]\vec{BA}[/Tex] and [Tex]\vec{BC}[/Tex]].

Vector [Tex]\vec{BA}[/Tex]:

[Tex]\vec{BA} = \vec{A} – \vec{B} = (1 – (-1), 2 – 0, 3 – 0) = (2, 2, 3)[/Tex]

Vector [Tex]\vec{BC}[/Tex]:

[Tex]\vec{BC} = \vec{C} – \vec{B} = (0 – (-1), 1 – 0, 2 – 0) = (1, 1, 2)[/Tex]

Dot product:

[Tex]\vec{BA} \cdot \vec{BC}[/Tex] = = (2 · 1) + (2 · 1) + (3 · 2) = 2 + 2 + 6 = 10

Magnitude of [Tex]\vec{BA}[/Tex]:

[Tex]\|\vec{BA}\| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}[/Tex]

Magnitude of [Tex]\vec{BC}[/Tex]:

[Tex]\|\vec{BC}\| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}[/Tex]

Now, plug these values into the formula for cosine:

[Tex]\cos(\theta) = \frac{10}{\sqrt{17} \cdot \sqrt{6}}[/Tex]

[Tex]\theta = \arccos\left(\frac{10}{\sqrt{17} \cdot \sqrt{6}}\right)[/Tex]

16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Direction vector AB:

AB = (2 – 1, 6 – 2, 3 – 7) = (1, 4, -4)

Direction vector BC:

BC = (3 – 2, 10 – 6, -1 – 3) = (1, 4, -4)

Since both direction vectors are the same, i.e., (1, 4, -4), the points A, B, and C are collinear.

17. Show that the vectors [Tex]\hat{2i} – \hat{j} + \hat{k}[/Tex], [Tex]\hat{i} – \hat{3j} – \hat{5k}[/Tex] and [Tex]\hat{3i} – \hat{4j} – \hat{4k}[/Tex] form the vertices of a right angled triangle

[Tex]|\hat{2i} – \hat{j} + \hat{k}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}[/Tex]

[Tex]|\hat{i} – \hat{3j} – \hat{5k}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}[/Tex]

[Tex]|\hat{3i} – \hat{4j} – \hat{4k}| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41}[/Tex]

√6² + √35² = √41²

6 + 35 = 41

41 = 41

Since the equation holds true, the vectors form the vertices of a right-angled triangle.

18. If [Tex]\vec{a}[/Tex] is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ[Tex]\vec{a}[/Tex] is unit vector if

(A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ|

For λa to be a unit vector, we want its magnitude to equal 1, so:

[Tex]| \lambda | |\vec{a}| = 1[/Tex]

Given [Tex]|\vec{a}| = a[/Tex] (as [Tex]\vec{a}[/Tex] is a nonzero vector of magnitude ‘a’), we have:

| λ | a = 1

To solve for λ, divide both sides by a:

| λ | = 1/a

Now, considering λ is a nonzero scalar, it can be either positive or negative, so |λ| is always positive.

Therefore, | λ | = 1/a is satisfied when a = 1/| λ |.

Thus, the correct answer is (D).

Read More:

• Vector Algebra
• Class 12 NCERT Solutions Chapter 10 – Vector Algebra Exercise 10.1
• Class 12 NCERT Solutions Chapter 10 – Vector Algebra Exercise 10.2
• Class 12 NCERT Solutions Chapter 10 – Vector Algebra Miscellaneous Exercise
• Class 12 Math NCERT solutions– Part- ii