Question

Let α→ = $2\^{i}-\^{j}+5\^{k} and b→= α\^{i}+β\^{j}+2\^{k}. If ((α→×b→)×\^{i}).\^{k}$=$\frac{ 23}{2}$,then $|b→×2\^{j}\|$is equal to

• A. 4
• B. 5
• C. √21
• D. √17

Answer 1

Answer: (B)

5

Answer 2

The correct answer is (B):
Given,
$α→=2\^{j}-\^{j}+5\^{k}$
and
$b→ = α\^{j}+β\^{j}+2\^{k}$
Also,
$((α→×b→)×i).\^{k} = \frac{23}{2}$
$⇒ ((α→.\^{i})b - (b→.\^{i}).α→).\^{k} = \frac{23}{2}$
⇒ $(2.b→-α.α→).\^{k} = \frac{23}{2}$
⇒ 2.2-5α = $\frac{23}{2}$
⇒ α = $\frac{-3}{2}$
Now, $|b→×2\^{j}| = |(α6\^{i}+β6\^{j}+2\^{k})×2\^{j|}$
$= |2α\^{k}+0-4\^{i}|$
= √4α2+16
= $√4(\frac{-3}{2})2+16$
= 5