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The algebra of vectors is a branch of mathematics that deals with vector quantities—objects defined by both magnitude and direction, unlike scalars, which have only magnitude. This field is essential in physics, engineering, and computer science for analyzing forces, velocities, and other vector quantities.
Solution:
We have,
Solution:
We have,
On squaring both sides,
Solution:
Given,
Let is a vector parallel to Therefore, for any scalar
Solution:
(i) We have, A = (4,-1) B = (1,3) Position Vector of A = Position Vector of B = Now, Therefore, (ii) We have, A = (-6,3) B = (-2,-5) Position Vector of A = Position Vector of B = Now, Therefore,
Solution:
We have,
A = (-1,3)
B = (-2,1)
Now,
Position Vector of
Position Vector of
Therefore,
Coordinate of the position vector
Solution:
Here, A = (-2,-1)
B = (3,0)
C = (1,-2)
Let us assume D be (x , y).
Computing Position Vector of AB, we have,
= Position Vector of B - Position Vector of A
Comparing LHS and RHS of both,
5 = 1-x
x = -4
And,
1 = -2-y
y = -3
So, coordinates of D = (-4,-3).
Solution:
Computing the position vectors of all the points we have,
Now,
Computing the final value after substituting the values,
Solution:
Given, Coordinate of A = (4,-1) Position vector of A = Position vector of Let coordinate of point B = (x, y) Position vector of B = Given that, Position vector of B - Position vector of A = \vec{a} Comparing the coefficients of LHS and RHS x - y = 5 x = 9 Also, y + 1 = 3 y = -1 So, coordinate of B = (9,-4)
Solution:
So, the two sides AB and AC of the triangle ABC are equal.
Therefore, ABC is an isosceles triangle.
Solution:
We have,
Let
Suppose is any vector parallel to
, where λ is any scalar.
Unit vector of
Therefore,
(i) P(3,2)
(ii) Q(-5,1)
(iii) R(-11,-9)
(iv) S(4,-3)
Solution:
(i) Given, P = (3,2)
Position vector of P =
Component of P along x-axis =
Component of P along y-axis =
(ii) Given, Q = (-5,1)
Position vector of Q =
Component of Q along x-axis =
Component of Q along y-axis =
(iii) Given, R = (-11,-9)
Position vector of R =
Component of R along x-axis =
Component of R along y-axis =
(iv) Given, S = (4,-3)
Position vector of S =
Component of S along x-axis =
Component of S along y-axis =
Related Article:
Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.6
Class 12 RD Sharma Solutions - Chapter 23 Algebra of Vectors - Exercise 23.4
Exercise 23.5 in Chapter 23 typically covers the topic of vector triple product. The vector triple product is a vector operation that involves three vectors and two cross product operations. It's usually denoted as a × (b × c), where a, b, and c are vectors.
The vector triple product is not associative: a × (b × c) ≠ (a × b) × c
The expansion formula for vector triple product is:
a × (b × c) = (a • c)b - (a • b)c
This formula is also known as Lagrange's formula or the BAC-CAB rule
The vector triple product can be used to solve various problems in physics and engineering
1). Given vectors a = 2i + 3j + k, b = i + 2j - k, and c = 3i - j + 2k, calculate a × (b × c).
2). Prove that a × (b × c) + b × (c × a) + c × (a × b) = 0 for any vectors a, b, and c.
3). If a • (b × c) = 0, what can you conclude about the vectors a, b, and c?
4). Show that (a × b) • (c × d) = (a • c)(b • d) - (a • d)(b • c).
5). Prove that a × (b × c) = (a • c)b - (a • b)c using the properties of dot and cross products.
6). If a, b, and c are mutually perpendicular unit vectors, evaluate a × (b × c).
7). Given that a • b = 2, b • c = 3, c • a = 4, and |a| = 2, |b| = 3, |c| = 4, calculate the scalar triple product a • (b × c).
8). Prove that a × (b × c) = (a × b) × c if and only if a, b, and c are coplanar.
9). If a × (b × c) = 0, what can you conclude about vectors a, b, and c?
10). Given vectors a = i - 2j + 3k, b = 2i + j - k, and c = -i + 3j + 2k, calculate (a × b) • (b × c).