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Mathematics Question on Vectors

If a⃗,b⃗,c⃗\vec{a} , \vec{b} , \vec{c} are mutually perpendicular vectors having magnitudes 1, 2, 3 respectively, then [a⃗+b⃗+c⃗  b⃗−a⃗−c⃗]=?[\vec{a} + \vec{b} + \vec{c} \, \, \vec{b} - \vec{a} - \vec{c}] = ?

A

0

B

6

C

12

D

18

Answer

12

Explanation

Solution

We have,
a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are mutually perpendicular vectors.
∴a⃗⋅b⃗=b⃗⋅c⃗=a⃗⋅c⃗=0\therefore \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{c}=0
and ∣a⃗∣=1,∣b⃗∣=2∣c⃗∣=3|\vec{a}|=1,|\vec{b}|=2|\vec{c}|=3
Now, [a⃗+b⃗+c⃗  b⃗−a⃗  c⃗][\vec{a}+\vec{b}+\vec{c} \,\,\vec{b}-\vec{a} \,\,\vec{c}]
=(a⃗+b⃗+c⃗)⋅((b⃗−a⃗)×c⃗)=(\vec{a}+\vec{b}+\vec{c}) \cdot((\vec{b}-\vec{a}) \times \vec{c})
=(a⃗+b⃗+c⃗)⋅[(b⃗×c⃗)−(a⃗×c⃗)]=(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{b} \times \vec{c})-(\vec{a} \times \vec{c})]
=a⃗⋅(b⃗×c⃗)−a⃗⋅(a⃗×c⃗)+b⃗⋅(b⃗×c⃗)−b⃗⋅(a⃗×c⃗)=\vec{a} \cdot(\vec{b} \times \vec{c})-\vec{a} \cdot(\vec{a} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{c})
+c⃗⋅(b⃗×c⃗)−c⃗⋅(a⃗×c⃗)+\vec{c} \cdot(\vec{b} \times \vec{c})-\vec{c} \cdot(\vec{a} \times \vec{c})
=[a⃗  b⃗  c⃗]−[a⃗  a⃗  c⃗]+[b⃗  b⃗  c⃗]−[b⃗  a⃗  c⃗]=[\vec{a} \,\,\vec{b}\,\, \vec{c}]-[\vec{a} \,\,\vec{a}\,\, \vec{c}]+[\vec{b} \,\,\vec{b} \,\,\vec{c}]-[\vec{b}\,\, \vec{a} \,\,\vec{c}]
+[c⃗ b⃗ c⃗]−[c⃗ a⃗ c⃗]+[\vec{c}\, \vec{b} \,\vec{c}]-[\vec{c}\, \vec{a} \,\vec{c}]
=[a⃗ b⃗ c⃗]−0+0+[a⃗ b⃗ c⃗]+0−0=[\vec{a} \,\vec{b} \,\vec{c}]-0+0+[\vec{a}\, \vec{b} \,\vec{c}]+0-0
=2[a⃗ b⃗ c⃗]=2[\vec{a}\, \vec{b} \,\vec{c}]
=2a⃗⋅(b⃗×c⃗)=2a⃗⋅(∣b⃗∣∣c⃗∣sin⁡π2n^)=2 \vec{a} \cdot(\vec{b} \times \vec{c})=2 \vec{a} \cdot\left(|\vec{b}||\vec{c}| \sin \frac{\pi}{2} \hat{n}\right)
=2a⃗⋅(2×3×1×n^)=12a⃗⋅n^=2 \vec{a} \cdot(2 \times 3 \times 1 \times \hat{n})=12 \vec{a} \cdot \hat{n}
=12∣a⃗∣∣n^∣cos⁡0∘=12×1×1×1=12=12|\vec{a}||\hat{n}| \cos 0^{\circ}=12 \times 1 \times 1 \times 1=12