Question

iii. Find a joint equation of the lines through the origin and perpendicular to the lines a x ^ 2 + 2xy * cot a - y ^ 2 = 0​

Answer 1

Step-by-step explanation:

Answer :

(2) √3, 120°

Explanation :

  The sum of vectors  $\vec{P} \ and \ \vec{Q}$  is given by,  

                       $\boxed{\bf R=\sqrt{P^2+Q^2+2PQcos\theta}}$

           Given,

         The sum of two unit vectors is also a unit vector,

               $\longrightarrow \ \ \ \ 1=\sqrt{1^2+1^2+2(1)(1)cos\theta} \\\\ \longrightarrow \ \ \ \ 1^2=1+1+2cos\theta \\\\ \longrightarrow \ \ \ \ 1=2+2cos\theta \\\\ \longrightarrow \ \ \ \ 1-2=2cos\theta \\\\ \longrightarrow \ \ \ \ 2cos\theta=-1 \\\\ \longrightarrow \ \ \ \ cos \theta=\frac{-1}{2} \\\\ \longrightarrow \ \ \ \ cos\theta=cos120^0 \\\\ \longrightarrow \ \ \ \ \boxed{\bf \theta=120^{\circ}}$  

The angle between the given two unit vectors = 120°      

  The magnitude of difference of two vectors is given by,

                       $\boxed{\bf S=\sqrt{P^2+Q^2-2PQcos\theta}}$

                $\longrightarrow \ \ \ \ S=\sqrt{1^2+1^2-2(1)(1)cos 120^{\circ}} \\\\ \longrightarrow \ \ \ \ S=\sqrt{1+1-2(\frac{-1}{2}) } \\\\ \longrightarrow \ \ \ \ S=\sqrt{2-(-1)} \\\\ \longrightarrow \ \ \ \ S=\sqrt{2+1} \\\\ \longrightarrow \ \ \ \ \boxed{\bf S=\sqrt{3} }$

The magnitude of the difference of the given two unit vectors is √3

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=> Unit vector :

  A vector of magnitude 1.

  Also known as direction vector.

  It has no units and dimensions.

  Unit vector = vector/vector's magnitude

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