Question 39 of 68intermediate🔧 ApplyNumerical4 marks

Find a unit vector perpendicular to each of the vector a⃗ + b⃗ and a⃗ - b⃗, where a⃗ = 3î + 2ĵ + 2k̂ and b⃗ = î + 2ĵ - 2k̂.

Correct Answer

The required unit vector is 23i^−23j^−13k^\frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3} \hat{k} (or its negative).

Exercise: EXERCISE 10.4 | Q: 2 | (Page 31)
For More Understanding

Explanation

Following the method in Example 23, we first compute the sum and difference vectors, then find their cross product which gives a vector perpendicular to both. Finally, we divide by the magnitude to get the unit vector.

Solution Steps

  1. Step 1: Find the vectors a⃗+b⃗\vec{a} + \vec{b} and a⃗−b⃗\vec{a} - \vec{b}.

  2. a⃗+b⃗\vec{a} + \vec{b} = (3i^+2j^+2k^\hat{i} + 2 \hat{j} + 2 \hat{k}) + (i^+2j^−2k^\hat{i} + 2 \hat{j} - 2 \hat{k}) = 4i^+4j^+0k^\hat{i} + 4 \hat{j} + 0 \hat{k}

  3. a⃗−b⃗\vec{a} - \vec{b} = (3i^+2j^+2k^\hat{i} + 2 \hat{j} + 2 \hat{k}) - (i^+2j^−2k^\hat{i} + 2 \hat{j} - 2 \hat{k}) = 2i^+0j^+4k^\hat{i} + 0 \hat{j} + 4 \hat{k}

  5. Step 2: Find the cross product (a⃗+b⃗\vec{a} + \vec{b}) × (a⃗−b⃗\vec{a} - \vec{b}).

  6. (a⃗+b⃗\vec{a} + \vec{b}) × (a⃗−b⃗\vec{a} - \vec{b}) = ∣i^j^k^440204∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}

  7. = i^\hat{i}(16 - 0) - j^\hat{j}(16 - 0) + k^\hat{k}(0 - 8)

  8. = 16i^−16j^−8k^\hat{i} - 16 \hat{j} - 8 \hat{k} (= c⃗\vec{c}, say)

  10. Step 3: Find the magnitude of c⃗\vec{c}.

  11. |c⃗\vec{c}| = 162+(−16)2+(−8)2=256+256+64=576=24\sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24

  13. Step 4: Find the required unit vector.

  14. Required unit vector = c⃗∣c⃗∣=16i^−16j^−8k^24=23i^−23j^−13k^\frac{\vec{c}}{|\vec{c}|} = \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3} \hat{k}

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