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Question: Let $\overline{a} = \hat{i} + \hat{j} + \hat{k}$, $\overline{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ are...

Let a‾=i^+j^+k^\overline{a} = \hat{i} + \hat{j} + \hat{k}, b‾=2i^−j^+3k^\overline{b} = 2\hat{i} - \hat{j} + 3\hat{k} are two vectors then a vector perpendicular to a‾+b‾\overline{a} + \overline{b} and a‾−b‾\overline{a} - \overline{b} and has magnitude 3263\sqrt{26} is

Answer

12\hat{i} - 3\hat{j} - 9\hat{k}

Explanation

Solution

  1. Calculate the sum and difference of vectors a‾\overline{a} and b‾\overline{b}: a‾+b‾=(i^+j^+k^)+(2i^−j^+3k^)=3i^+4k^\overline{a} + \overline{b} = (\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = 3\hat{i} + 4\hat{k} a‾−b‾=(i^+j^+k^)−(2i^−j^+3k^)=−i^+2j^−2k^\overline{a} - \overline{b} = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) = -\hat{i} + 2\hat{j} - 2\hat{k}

  2. A vector perpendicular to two vectors u‾\overline{u} and v‾\overline{v} is given by their cross product u‾×v‾\overline{u} \times \overline{v}. The vector perpendicular to (a‾+b‾)(\overline{a} + \overline{b}) and (a‾−b‾)(\overline{a} - \overline{b}) is proportional to (a‾+b‾)×(a‾−b‾)(\overline{a} + \overline{b}) \times (\overline{a} - \overline{b}). Using vector properties, (a‾+b‾)×(a‾−b‾)=a‾×a‾−a‾×b‾+b‾×a‾−b‾×b‾=0‾−(a‾×b‾)−(a‾×b‾)−0‾=−2(a‾×b‾)(\overline{a} + \overline{b}) \times (\overline{a} - \overline{b}) = \overline{a} \times \overline{a} - \overline{a} \times \overline{b} + \overline{b} \times \overline{a} - \overline{b} \times \overline{b} = \overline{0} - (\overline{a} \times \overline{b}) - (\overline{a} \times \overline{b}) - \overline{0} = -2(\overline{a} \times \overline{b}). Thus, the required vector is in the direction of a‾×b‾\overline{a} \times \overline{b}.

  3. Calculate the cross product a‾×b‾\overline{a} \times \overline{b}: a‾×b‾=∣i^j^k^1112−13∣=(3−(−1))i^−(3−2)j^+(−1−2)k^=4i^−j^−3k^\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{vmatrix} = (3 - (-1))\hat{i} - (3 - 2)\hat{j} + (-1 - 2)\hat{k} = 4\hat{i} - \hat{j} - 3\hat{k}.

  4. Calculate the magnitude of a‾×b‾\overline{a} \times \overline{b}: ∣a‾×b‾∣=∣4i^−j^−3k^∣=42+(−1)2+(−3)2=16+1+9=26|\overline{a} \times \overline{b}| = |4\hat{i} - \hat{j} - 3\hat{k}| = \sqrt{4^2 + (-1)^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26}.

  5. The required vector, let's call it V‾\overline{V}, must have a magnitude of 3263\sqrt{26} and be in the direction of a‾×b‾\overline{a} \times \overline{b}. So, V‾=k(a‾×b‾)\overline{V} = k (\overline{a} \times \overline{b}) for some scalar kk. The magnitude is ∣V‾∣=∣k∣∣a‾×b‾∣|\overline{V}| = |k| |\overline{a} \times \overline{b}|. Given ∣V‾∣=326|\overline{V}| = 3\sqrt{26}, we have 326=∣k∣263\sqrt{26} = |k| \sqrt{26}. This implies ∣k∣=3|k| = 3, so k=3k = 3 or k=−3k = -3.

  6. For k=3k=3: V‾=3(4i^−j^−3k^)=12i^−3j^−9k^\overline{V} = 3(4\hat{i} - \hat{j} - 3\hat{k}) = 12\hat{i} - 3\hat{j} - 9\hat{k}. Magnitude: ∣12i^−3j^−9k^∣=122+(−3)2+(−9)2=144+9+81=234=9×26=326|12\hat{i} - 3\hat{j} - 9\hat{k}| = \sqrt{12^2 + (-3)^2 + (-9)^2} = \sqrt{144 + 9 + 81} = \sqrt{234} = \sqrt{9 \times 26} = 3\sqrt{26}.

  7. For k=−3k=-3: V‾=−3(4i^−j^−3k^)=−12i^+3j^+9k^\overline{V} = -3(4\hat{i} - \hat{j} - 3\hat{k}) = -12\hat{i} + 3\hat{j} + 9\hat{k}. Magnitude: ∣−12i^+3j^+9k^∣=(−12)2+32+92=144+9+81=234=326|-12\hat{i} + 3\hat{j} + 9\hat{k}| = \sqrt{(-12)^2 + 3^2 + 9^2} = \sqrt{144 + 9 + 81} = \sqrt{234} = 3\sqrt{26}.

Both vectors satisfy the conditions. The question asks for "a vector". One such vector is 12i^−3j^−9k^12\hat{i} - 3\hat{j} - 9\hat{k}.