Question 40 of 68beginner🔧 ApplyNumerical1 mark

Find the projection of the vector î - ĵ on the vector î + ĵ.

Correct Answer

0

Exercise: EXERCISE 10.3 | Q: 3 | (Chapter: 24)
For More Understanding

Explanation

The question asks for the projection of one vector onto another. According to the 'Observations' section in the provided context, specifically Observation 2, the projection of a vector a⃗\vec{a} on vector b⃗\vec{b} is given by the formula 1∣b⃗∣(a⃗⋅b⃗)\frac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b}). We identify a⃗=i^−j^\vec{a} = \hat{i} - \hat{j} and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}. The dot product a⃗⋅b⃗\vec{a} \cdot \vec{b} is calculated as 0, which results in the projection being 0. This is consistent with the geometric interpretation that if the dot product is zero, the vectors are perpendicular, and the projection of one onto the other is zero.

Solution Steps

  1. Step 1: Identify the vectors Let a⃗=i^−j^\vec{a} = \hat{i} - \hat{j} and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}.

  2. Step 2: Calculate the dot product (a⃗⋅b⃗\vec{a} \cdot \vec{b}) Using the property a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 provided in the context: a⃗⋅b⃗=(1)(1)+(−1)(1)+(0)(0)=1−1=0\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) + (0)(0) = 1 - 1 = 0.

  3. Step 3: Calculate the magnitude of vector b⃗\vec{b} (∣b⃗∣|\vec{b}|) ∣b⃗∣=(1)2+(1)2=1+1=2|\vec{b}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}.

  4. Step 4: Apply the projection formula Using the formula from Observation 2: Projection of a⃗\vec{a} on b⃗=1∣b⃗∣(a⃗⋅b⃗)\vec{b} = \frac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b}). Projection = 12(0)=0\frac{1}{\sqrt{2}} (0) = 0.

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