| 1. \[(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=\vec{0}\] |
| 2. \[(\vec{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \vec{d})=\vec{0}\] |
| Select the correct answer using the code given below: |
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 nor 2
Correct Answer: C
Solution :
[c] \[\left( \overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{d}}\, \right)\times \left( \overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{c}}\, \right)\] \[=\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{d}}\,\times \overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{d}}\,\times \overset{\to }{\mathop{c}}\,\] \[=\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,-\overset{\to }{\mathop{d}}\,\times \overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{d}}\,-\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,\] \[=-\overset{\to }{\mathop{d}}\,\times \overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{d}}\,\times \overset{\to }{\mathop{b}}\,\] \[=0\] Again \[(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)=(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,)\] given \[\Rightarrow (\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)\times (\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,)=(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,)\times (\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{d}}\,)=0\] \[\left( as\,\,\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{a}}\,=0 \right)\] So both (1) and (2) are correct.
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