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Exercise 10.2 · Q6

Q.Find the sum of the vectors a⃗=i^−2j^+k^\vec{a} = \hat{i} - 2\hat{j} + \hat{k}, b⃗=−2i^+4j^+5k^\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k} and c⃗=i^−6j^−7k^\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}.

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Vector addition is done component-wise: add the i^\hat{i}, j^\hat{j}, and k^\hat{k} coefficients separately. The sum is − 4j^−k^-\ 4\hat{j} - \hat{k}.

The idea is simple: when you add vectors, you combine their effects along each direction independently. Think of it like adding apples to apples and oranges to oranges — the i^\hat{i} parts only combine with other i^\hat{i} parts, and so on. This works because the unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k} are mutually perpendicular and form a basis for 3D space.

Let’s go through it step by step.

  1. Identify the components of each vector.

    a⃗=1i^−2j^+1k^\vec{a} = 1\hat{i} - 2\hat{j} + 1\hat{k}

    b⃗=−2i^+4j^+5k^\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}

    c⃗=1i^−6j^−7k^\vec{c} = 1\hat{i} - 6\hat{j} - 7\hat{k}

  2. Add the i^\hat{i}-components.

    1+(−2)+1=01 + (-2) + 1 = 0

    So the i^\hat{i}-component of the sum is 0i^0\hat{i} — it cancels out completely.

  3. Add the j^\hat{j}-components.

    −2+4+(−6)=−4-2 + 4 + (-6) = -4

    So the j^\hat{j}-component is −4j^-4\hat{j}.

  4. Add the k^\hat{k}-components.

    1+5+(−7)=−11 + 5 + (-7) = -1

    So the k^\hat{k}-component is −k^-\hat{k}.

  5. Write the resultant vector. …

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