Question

Assertion :If $i\times [(a-j)\times i]+j\times [(a-k)\times j]+k[(a-i)\times k]=0$, then $a=\frac{1}{2}(i+j+k)$. Reason: $(a.i)i+(a.j)j+(a.k)k=a$ for any vector $a$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We have,

$i\times [(a-j)\times i]=(i.i)(a-j)-[i.(a-j)]i=a-j-(i.a)i$

$j\times [(a-k)\times j]=(j.j)(a-k)-\left[(j.(a-k))\right]j=a-k-(j.a)j$

$k\times [(a-i)\times k]=(k.k)(a-i)-\left[(k.(a-i))\right]k=a-i-(k.a)k$

According to the given condition, we have

$i\times [(a-j)\times i]+j\times [(a-k)\times j]+k\times [(a-i)\times k]=0$

i.e., $3a-(i+j+k)-[(a.i)i+(a.j)j+(a.k)k]=0$

i.e., $3a-a=i+j+k$

i.e., $a=\frac{1}{2}(i+j+k).$