pls explain it how this formula derived. Class XI
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Here , you have to use ' dot product of two vectors' .
As you know,
Dot product of two vectors A and B = |A|.|B|cos {theta}
where |A| and |B| are the magnitude of A and B . And {theta} is the angle Between A and B.
{alpha} is angle between A and positive x-axis.
so, dot product of A and x-axis.
Here,
A = xi + yj +zk
x-axis = i
A.i = |A|.|i|cos {alpha}
we know,
|A| = sqrt {x^2 +y^2 +z^2}
|i| = sqrt {1^2} = 1 put it above .
(Xi + yj + zk).(i +0j +0k)= sqrt {x^2 +y^2+z^2}.1cos {alpha}
x.1 + y.0 + z.0 = sqrt {x^2 + y^2 + z^2} cos{alpha}
cos {alpha} = x/sqrt {x^2+y^2+z^2}
Similarly you can deprive
Cos {beta} = y/sqrt {x^2+y^2+z^2}
Cos {gama}= z/sqrt {x^2+y^2+z^2}
As you know,
Dot product of two vectors A and B = |A|.|B|cos {theta}
where |A| and |B| are the magnitude of A and B . And {theta} is the angle Between A and B.
{alpha} is angle between A and positive x-axis.
so, dot product of A and x-axis.
Here,
A = xi + yj +zk
x-axis = i
A.i = |A|.|i|cos {alpha}
we know,
|A| = sqrt {x^2 +y^2 +z^2}
|i| = sqrt {1^2} = 1 put it above .
(Xi + yj + zk).(i +0j +0k)= sqrt {x^2 +y^2+z^2}.1cos {alpha}
x.1 + y.0 + z.0 = sqrt {x^2 + y^2 + z^2} cos{alpha}
cos {alpha} = x/sqrt {x^2+y^2+z^2}
Similarly you can deprive
Cos {beta} = y/sqrt {x^2+y^2+z^2}
Cos {gama}= z/sqrt {x^2+y^2+z^2}
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