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Find a unit vector perpendicular to the plane containing vectors P = 3i − 2j + 4k and Q = i + j − 2k.

Mathematics

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Answer

n^=(B) ±15(2j⃗+k⃗)\hat{n} = \text{(B) } \pm \frac{1}{\sqrt{5}}(2\vec{j} + \vec{k})

Step-by-step solution

3 steps

To find a unit vector perpendicular to the plane containing two vectors, we first calculate their cross product, and then divide the resulting vector by its magnitude.

Given vectors:
P⃗=3i⃗−2j⃗+4k⃗\vec{P} = 3\vec{i} - 2\vec{j} + 4\vec{k}
Q⃗=i⃗+j⃗−2k⃗\vec{Q} = \vec{i} + \vec{j} - 2\vec{k}

  1. Calculate the cross product N⃗=P⃗×Q⃗\vec{N} = \vec{P} \times \vec{Q}.

    N⃗=∣i⃗j⃗k⃗3−2411−2∣\vec{N} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & -2 & 4 \\ 1 & 1 & -2 \end{vmatrix} N⃗=i⃗((−2)(−2)−(4)(1))−j⃗((3)(−2)−(4)(1))+k⃗((3)(1)−(−2)(1))\vec{N} = \vec{i}((-2)(-2) - (4)(1)) - \vec{j}((3)(-2) - (4)(1)) + \vec{k}((3)(1) - (-2)(1)) N⃗=i⃗(4−4)−j⃗(−6−4)+k⃗(3−(−2))\vec{N} = \vec{i}(4 - 4) - \vec{j}(-6 - 4) + \vec{k}(3 - (-2)) N⃗=0i⃗−j⃗(−10)+k⃗(5)\vec{N} = 0\vec{i} - \vec{j}(-10) + \vec{k}(5) N⃗=10j⃗+5k⃗\vec{N} = 10\vec{j} + 5\vec{k}
  2. Calculate the magnitude of N⃗\vec{N}.

    ∣N⃗∣=(0)2+(10)2+(5)2|\vec{N}| = \sqrt{(0)^2 + (10)^2 + (5)^2} ∣N⃗∣=0+100+25|\vec{N}| = \sqrt{0 + 100 + 25} ∣N⃗∣=125|\vec{N}| = \sqrt{125} ∣N⃗∣=25×5|\vec{N}| = \sqrt{25 \times 5} ∣N⃗∣=55|\vec{N}| = 5\sqrt{5}
  3. Find the unit vector n^\hat{n}.

    A unit vector perpendicular to the plane can point in two opposite directions, so we include ±\pm.

    n^=±N⃗∣N⃗∣=±10j⃗+5k⃗55\hat{n} = \pm \frac{\vec{N}}{|\vec{N}|} = \pm \frac{10\vec{j} + 5\vec{k}}{5\sqrt{5}} n^=±(1055j⃗+555k⃗)\hat{n} = \pm \left( \frac{10}{5\sqrt{5}}\vec{j} + \frac{5}{5\sqrt{5}}\vec{k} \right) n^=±(25j⃗+15k⃗)\hat{n} = \pm \left( \frac{2}{\sqrt{5}}\vec{j} + \frac{1}{\sqrt{5}}\vec{k} \right) n^=(B) ±15(2j⃗+k⃗)\hat{n} = \boxed{\text{(B) } \pm \frac{1}{\sqrt{5}}(2\vec{j} + \vec{k})}
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