Answer to Question #245546 in Analytic Geometry for moe

Question

Consider the vectors $\mathbf{a}=(-k, 1+k, k(1+k))$ and $\mathbf{b}=(1+k, k(1+k), -k)$ where k is some constant. Which of the following is/are correct?

1. a and b are perpendicular vectors for any k
2. ∣a∣=∣b∣.
3. a and b are unit vectors for k=1
4. The values of k for which ∣a∣=1 are 0 and -1
5. The two vectors above are parallel.

Answer 1

1.

If two vectors are perpendicular then $\overrightarrow{A}.\overrightarrow{B}=0$

$[-k\widehat{i}+(i+k)\widehat{j}+9k+k^2)\widehat{k}].[(1+k)\widehat{i}+(k+k^2)\widehat{j}-k\widehat{k}=0$

$-k(1+k)+(1+k)(k+k^2)+(k+k^2)(-k)=0$

$-k-k^2+k+k^2+k^2+k^3-k^2-k^3=0\\0=0$

option 1 is correct

2.

$|\overrightarrow{A}|=\sqrt{(-k)^2+(1+k)^2+(k+k^2)}$

$\\=\sqrt{k^2+1+k^2+2k+k^2+k^4+2k^3}$

$=\sqrt{k^4+2k^3+3k^2+2k+1}$

$\\|\overrightarrow{B}|=\sqrt{(1+k)^2+(k+k^2)+(-k)^2}$

$\\=\sqrt{1+k^2+2k+k^2+k^4+2k^3+k^2}$

$\\\sqrt{k^4+2k^3+3k^2+2k+1}$

$\\|\overrightarrow{A}|=|\overrightarrow{B}|$

option 2 is correct

3.

if $\overrightarrow{A} \& \overrightarrow{B}$ is a unit vector

$|\overrightarrow{A}|=1,\space |\overrightarrow{B}|=1$

$k^4+2k^3+3k^2+2k+1=1\\k(k^3+2k^2+3k+2)=0................(1)\\for\space k=1$

equation (1) not satisfied so option (3) is not correct

4.

$|\overrightarrow{A}|=1$

by equation (1)

$k((k^3+2k^2+3k+2)=0\\k=0\\k^3+2k^2+3k+2=0\\put\space k=-1\\-1+2-3+2=0\\0=0$

option 4 is correct

5.

vector is parallel if

$-k=d(1+k)\space \&\space 1+k=d(k(1+k)\\\frac{-k}{1+k}=d,\space \frac{1}{k}=d\space \&\space k(1+k)=d(-k),\space d=(-1+k)$

Therefore the vectors are not parallel