How is the area of a parallelogram calculated?

The area of a parallelogram is found by multiplying the base by the height.

How is the angle between vectors A and B calculated?

The angle between vectors A and B can be determined using the dot product formula.

What is the cosine of 150 degrees?
-√3/2
How is the area of a parallelogram calculated?
The area of a parallelogram is found by multiplying the base by the height.
What is the magnitude of vector B?
3
How is the angle between vectors A and B calculated?
The angle between vectors A and B can be determined using the dot product formula.
What is the value of x in the vector equation x² + x² + x² = 1?
x = ±1/√3

00:00

Dot product is commutative: a dot b = z, b dot a = z
Vectors chapter discussion with 30 important questions
Start solving KCT exam problems from April 21st to February
Angle between vectors a and b is 180°
Solution to a + 2b + 3c = 0 is 6b x c
Vector makes angles of 150° and 60° with x and y axes respectively
Cosine of 150° is -√3/2, cosine of 60° is 1/2
Angle between vectors a and b is 2π/3
Magnitude of vector b is 3
Diagonals of a parallelogram are calculated using vector addition: D1 = a + b, D2 = a - b

22:50

Magnitude of vector D1 is root 3, and magnitude of vector D2 is root 2, with the correct answer being the first option.
A and B are perpendicular, leading to the dot product of A and B being zero, and the need to find the value of M.
The area of a parallelogram with adjacent sides A and B is the square root of 3.
A and B, inclined at an angle of Pi by 3, result in the value of A plus B being greater than 1.
The area of a triangle formed by vectors A and B, with sides of 1 and 2, is 15 by 4 square units.
The magnitude of vector B is root 7, given A dot B equals magnitude of B squared and magnitude of A minus B equals root 7.
The angle between vectors A and B, with A dot B equal to magnitude of B squared, is 30 degrees.
The angle between A and B, with 3 in root 3 times A minus B as a unit vector, is 60 degrees.
The angle between A and B, with A plus B minus C resulting in cos 1 being 1 by 2, is Pi by 3.
Vector A, B, and C, with magnitudes 3, 4, and 5 respectively, are perpendicular to the sum of the remaining vectors, leading to the sum of their squares being 50.

41:31

The equation c² = 50 implies that A + B + C equals the square root of 25 multiplied by 2, resulting in 5 times the square root of 2.
The value of x in a vector equation is found by squaring x² + x² + x² = 1, leading to x = ±1/√3.
The direction cosines of vector A are cos Alpha, cos Beta, and cos Gamma, with the sum of cos Beta and cos Gamma equaling 1.
The cosines of 2 Alpha, 2 Beta, and 2 Gamma are calculated, resulting in cos 2 Alpha = -1.
For orthogonal vectors A and B, the value of Lambda is determined to be -5/2.
Mutually perpendicular unit vectors A and B lead to a final answer of 3 for a specific calculation.
The values of Lambda and Mu are found to be 1/2 and 7/4, respectively, for orthogonal vectors A and B.
The projection of vector A on B results in a magnitude of 4.
A unit vector perpendicular to the plane containing two given vectors is calculated as ±(I - J + K) / √3.
The sine of Theta/2 is derived as (1 - cos Theta) / 2, leading to a final answer of (A - B) / 2√2 for given unit vectors A and B.

01:01:11

Vector A + Vector B is equal to 2I + 4J + 6K
Magnitude of Vector R is the square root of 1 + 4 + 9, which equals root 14
The correct answer is option C
Magnitude of Vector B is 2
The correct answer is option B
The area of the quadrilateral ABCD is 9 square units
Vector A is 1/3I + 1/3J + 1/3K
The position of Vector B on Vector A is 11/3
The correct answer is option B
Magnitude of Vector B is 3 times the magnitude of Vector A
The correct answer is the fourth option in the question
The session covered 30 questions from the Super 30 problem set in preparation for KCT exams.