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For any vector a⃗\vec{a}, the value of ∣a⃗×i^∣2+∣a⃗×j^∣2+∣a⃗×k^∣2|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 is equal to:

Solution

✅ Correct Option: 1

Let a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}


For a⃗×i^\vec{a} \times \hat{i}:

a⃗×i^=(a1i^+a2j^+a3k^)×i^\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i}

Using i^×i^=0\hat{i} \times \hat{i} = 0, j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, and k^×i^=j^\hat{k} \times \hat{i} = \hat{j}:

a⃗×i^=a3j^−a2k^\vec{a} \times \hat{i} = a_3\hat{j} - a_2\hat{k}

∣a⃗×i^∣2=a32+a22|\vec{a} \times \hat{i}|^2 = a_3^2 + a_2^2


For a⃗×j^\vec{a} \times \hat{j}:

a⃗×j^=(a1i^+a2j^+a3k^)×j^\vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j}

Using i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×j^=0\hat{j} \times \hat{j} = 0, and k^×j^=−i^\hat{k} \times \hat{j} = -\hat{i}:

a⃗×j^=−a3i^+a1k^\vec{a} \times \hat{j} = -a_3\hat{i} + a_1\hat{k}

∣a⃗×j^∣2=a32+a12|\vec{a} \times \hat{j}|^2 = a_3^2 + a_1^2


For a⃗×k^\vec{a} \times \hat{k}:

a⃗×k^=(a1i^+a2j^+a3k^)×k^\vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k}

Using i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, and k^×k^=0\hat{k} \times \hat{k} = 0:

a⃗×k^=a2i^−a1j^\vec{a} \times \hat{k} = a_2\hat{i} - a_1\hat{j}

∣a⃗×k^∣2=a22+a12|\vec{a} \times \hat{k}|^2 = a_2^2 + a_1^2


∣a⃗×i^∣2+∣a⃗×j^∣2+∣a⃗×k^∣2|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2

=(a22+a32)+(a12+a32)+(a12+a22)= (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2)

=2a12+2a22+2a32= 2a_1^2 + 2a_2^2 + 2a_3^2

=2(a12+a22+a32)= 2(a_1^2 + a_2^2 + a_3^2)

=2∣a⃗∣2= 2|\vec{a}|^2

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