Let a and b are two non-zero vectors perpendicular to each other and |a| = |b|. If x x b = a, then the angle between the vectors (a + b + (a x b)) and a is equal to :

<p>Given vectors a? and b? such that |a? | = |b? | and a? ⋅ b? = 0 (they are orthogonal).<br>The problem implies |a? |=|b? |=1.<br>Let c? = a? + b? + a? x b? <br>To find the magnitude of c? , we calculate |c? |²:<br>|c? |² = c? ⋅ c? = (a? + b? + a? x b? ) ⋅ (a? + b? + a? x b? ).<br>This expands to |a? |² + |b? |² + |a? x b? |² because all other dot products are zero (e.g., a? ⋅ b? = 0, a? ⋅ (a? x b? ) = 0).<br>|a? x b? |² = (|a? |b? |sin (90°)² = |a? |²|b? |².<br>So, |c? |² = |a? |² + |b? |² + |a? |²|b? |² = 1² + 1² + 1²*1² = 3.<br>∴ |c? | = √3.<br>To find the angle θ between c? and a? , we compute their dot product:<br>c? ⋅ a? = (a? + b? + a? x b? ) ⋅ a? = a? ⋅ a? + b? ⋅ a? + (a? x b? ) ⋅ a? <br>c? ⋅ a? = |a? |² + 0 + 0 = 1.<br>Now, cos θ = (c? ⋅ a? ) / (|c? |a? |) = 1 / (√3 * 1) = 1/√3.<br>⇒ θ = cos? ¹ (1/√3).</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment