Question 29 of 68intermediate🔧 ApplyNumerical2 marks

Find |x⃗|, if for a unit vector a⃗, (x⃗ - a⃗) ⋅ (x⃗ + a⃗) = 12.

Correct Answer

Step 1: Expand the given expression using the distributive property of the scalar product.

(x⃗−a⃗)⋅(x⃗+a⃗)=x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a}

Since x⃗⋅a⃗=a⃗⋅x⃗\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x} and x⃗⋅x⃗=∣x⃗∣2\vec{x} \cdot \vec{x} = |\vec{x}|^2 and a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2, this simplifies to:

∣x⃗∣2−∣a⃗∣2|\vec{x}|^2 - |\vec{a}|^2

Step 2: Substitute the known values into the equation. We are given that a⃗\vec{a} is a unit vector, so ∣a⃗∣=1|\vec{a}| = 1. We are also given that the expression equals 12. Therefore, ∣x⃗∣2−(1)2=12|\vec{x}|^2 - (1)^2 = 12

Step 3: Solve for ∣x⃗∣|\vec{x}|.

∣x⃗∣2−1=12|\vec{x}|^2 - 1 = 12 ∣x⃗∣2=13|\vec{x}|^2 = 13

Taking the square root, we get ∣x⃗∣=13|\vec{x}| = \sqrt{13}.

Exercise: EXERCISE 10.3 | Q: 9 | (Page 25)
For More Understanding

Explanation

This question is solved by expanding the dot product expression, a technique implied in similar problems like Question 6 provided in the context. The expansion follows the pattern (a−b)⋅(a+b)=∣a∣2−∣b∣2(a-b)\cdot(a+b) = |a|^2 - |b|^2. Since a⃗\vec{a} is defined as a unit vector in the question, its magnitude ∣a⃗∣|\vec{a}| is 1. Substituting these values allows for the direct calculation of ∣x⃗∣|\vec{x}|.

Solution Steps

  1. Step 1: Expand the dot product: (x⃗−a⃗)⋅(x⃗+a⃗)=∣x⃗∣2−∣a⃗∣2(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = |\vec{x}|^2 - |\vec{a}|^2.

  2. Step 2: Substitute ∣a⃗∣=1|\vec{a}| = 1 and equate to 12: ∣x⃗∣2−1=12|\vec{x}|^2 - 1 = 12.

  3. Step 3: Solve for ∣x⃗∣|\vec{x}|: ∣x⃗∣2=13  ⟹  ∣x⃗∣=13|\vec{x}|^2 = 13 \implies |\vec{x}| = \sqrt{13}.

PreviousNext