Question

In triangle ABC prove that tan(B-C)/2=b-c/b+c.cotA/2​

Answer 1

Question:

$\footnotesize{ \sf Three \: vectors \: A , B \: and \: C \: add \: upto \: zero. } \footnotesize{ \: \sf Find \: which \: is \: false }$

• $\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
• $\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
• $\footnotesize{ \sf C) \: If \: A,B ,C \: define \: a \: plane \: (A \times B)\times C \:is \: in \: that \: plane.}$
• $\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .}$

Things to know:

• Vector product of a vector with itself is always 0, $\vec{A} \times \vec{A} = A^2 sin \theta \implies A^2 sin^\circ \implies 0$
• Similarly, the vector product of two parallel vector is always 0 as the angle b/w them will be 0.
• $\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A})$
• $\vec{A} \times \vec{B}$ is perpendicular to both $\vec{A} and B$. This can be proved with right hand thumb rule. The cross product of two vectors is always directed perpendicularly outwards to the plane and hence it is parallel to its operands.

Solution:

$\tt Given~that~ \vec{A} + \vec{B} + \vec{C} =0 \\\\ \tt Taking~cross~product~of~ \vec{B} ~on~both~sides. We have, : \\\\ \tt \implies \vec{B} \times ( \vec{A} + \vec{B} + \vec{C}) = 0 \times \vec{B} \\\\ \tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{B} + \vec{B} \times \vec{C} = 0 \\\\\tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{C} = 0 \\\\ \tt\implies \vec{B} \times \vec{A} = - (\vec{B} \times \vec{C}) \\\\ \tt \implies \vec{B} \times \vec{A} = \vec{C} \times \vec{B}$

It can also be written as $\vec{A} \times \vec{B} = \vec{B} \times \vec{C} - - - - - [i]$

Now, come to the first option.

$\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$

From [i] we have, $\vec{A} \times \vec{B} = \vec{B} \times \vec{C}$

Taking vector product of C on both sides, we have:

$(\vec{A} \times \vec{B}) \times \vec{C} =( \vec{B} \times \vec{C}) \times \vec{C}$

$(\vec{A} \times \vec{B}) \times \vec{C}$ can only be zero if $\vec{B} ~and~\vec{C}$ are parallel I.e their vector product is zero

Hence, it is true.

Now, come to 2nd option.

$\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$

Similarly, Taking dot product of C vector on both sides in [i], we have :

$\tt (\vec{A} \times \vec{B}). \vec{C} =( \vec{B} \times \vec{C}). \vec{C}$

$(\vec{B} \times \vec{C}). \vec{C}$ will always be zero. It doesn't require $\vec{A} and \vec{B}$ to be parallel. It will gonna be zero in any case and so will be $(\vec{A} \times \vec{B}).\vec{C}$

Hence, Option B is incorrect as $(\vec{A} \times \vec{B}).\vec{C}$ can be zero even when B and C are not parallel.

Now come to 4th option.

$\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .}$

$( \vec{A} \times \vec{B}). \vec{C} \implies (|A| |B| sin \theta). \vec{C} \implies |(|A| |B| sin \theta)|~|C| cos \theta$ We can bring this $|(|A| |B| sin \theta)|~|C| cos \theta$ to |A| |B| |C| only when the angle b/w A and B is 90° as sin 90=1.

And, it is already given that, A+B+C=0, that means they are the sides of a triangle. And, hence they will satisfy, the equation $A^2+B^2=C^2$