Question

If p=i-2k and q=-3i+j+k find angle between p+q and p-q

Answer 1

P+q=4i+j-k
P-q= 4i-j-3k
CosA= (16-1+3)/square root of 18*26
18/_|18*26 so3/_| 13

Answer 2

Given:

$P=i-2k$

$Q=-3i+j+k$

To find:

The angle between $P+Q$ and $P-Q$.

Solution:

We have, $P=i-2k$ and $Q=-3i+j+k$ .

Hence,

$P+Q=(i-2k)+(-3i+j+k)$

    $A$    $=-2i+j-k$           $;$ and

$P-Q=i-2k-(-3i+j+k)$

           $=i-2k+3i-j-k$

    $B$    $=4i-j-3k$

We know, for any vectors $A$ and $B$,

$cos\alpha =\frac{vec(A).vec(B)}{|vec(A)||vec(B)|}$

Hence,

$cos\alpha =\frac{(-2i+j-k).(4i-j-3k)}{|-2i+j-k||4i-j-k|}$

$cos\alpha =\frac{-8-1+3}{\sqrt{(-2)^{2}+ (1)^{2}+ (-1)^{2} } \sqrt{(4)^{2}+ (-1)^{2} +(-1)^{2} } }$

$cos\alpha =\frac{-6}{\sqrt{4+1+1} \sqrt{(16+1+1)} }$

$cos\alpha =\frac{-6}{\sqrt{6} \sqrt{18} }$

$cos\alpha =\frac{-6}{6\sqrt{3} }$

$cos\alpha =\frac{-1}{\sqrt{3} }$

$\alpha =cos^{-1}(\frac{-1}{\sqrt{3} } )$

Final answer:

Hence, the angle between $P+Q$ and $P-Q$  is $cos^{-1}(\frac{-1}{\sqrt{3} } )$.