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Match List - I with List - II.

List - I (Two given vector)List - II (Projection of a⃗\vec{a} on b⃗\vec{b})
(A)a⃗=i^−j^\vec{a} = \hat{i} - \hat{j}, b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}(I)25\frac{2}{\sqrt{5}}
(B)a⃗=i^+j^\vec{a} = \hat{i} + \hat{j}, b⃗=2i^−k^\vec{b} = 2\hat{i} - \hat{k}(II)0
(C)a⃗=j^+k^\vec{a} = \hat{j} + \hat{k}, b⃗=i^+k^\vec{b} = \hat{i} + \hat{k}(III)2\sqrt{2}
(D)a⃗=2i^+3j^\vec{a} = 2\hat{i} + 3\hat{j}, b⃗=i^−k^\vec{b} = \hat{i} - \hat{k}(IV)12\frac{1}{\sqrt{2}}

Choose the correct answer from the options given below :

Solution

✅ Correct Option: 1

Projection of a⃗\vec{a} on b⃗\vec{b} is a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.

(A) a⃗⋅b⃗=1−1=0\vec{a} \cdot \vec{b} = 1-1 = 0. Proj =0= 0 (II).

(B) a⃗⋅b⃗=2\vec{a} \cdot \vec{b} = 2, ∣b⃗∣=5|\vec{b}| = \sqrt{5}. Proj =25= \frac{2}{\sqrt{5}} (I).

(C) a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1, ∣b⃗∣=2|\vec{b}| = \sqrt{2}. Proj =12= \frac{1}{\sqrt{2}} (IV).

(D) a⃗⋅b⃗=2\vec{a} \cdot \vec{b} = 2, ∣b⃗∣=2|\vec{b}| = \sqrt{2}. Proj =2= \sqrt{2} (III).

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