Answer to Question #226834 in Linear Algebra for Sarita bartwal

Question

a. Find a unit vector in R^3 that is orthogonal to (1,2,1) and (1,-1,2).
b. If T: R^5 to R^3 is a linear transformation, then there is u

Answer 1

$\text{The vector orthogonal to both (1, 2, 1) and (1, -1, 2) is the vector cross} \\\text{product between (1,2,1) and (1,-1,2)} \\\begin{vmatrix} i & j & k\\ 1 & 2 & 1 \\ 1 & -1 & 2 \end{vmatrix} \\=i\begin{vmatrix} 2 & 1\\ -1 & 2 \end{vmatrix} -j\begin{vmatrix} 1 & 1\\ 1 & 2 \end{vmatrix} + k\begin{vmatrix} 1 & 2\\ 1 & -1 \end{vmatrix}=5i-j-3k \\\text{The unit vector is} \\=\frac{5i-j-3k}{\sqrt{5^2+(-1)^2+(-3)^2}}=\frac{5i-j-3k}{\sqrt{35}}$$\text{The vector orthogonal to both (1, 2, 1) and (1, -1, 2) is the vector cross} \\\text{product between (1,2,1) and (1,-1,2)} \\\begin{vmatrix} i & j & k\\ 1 & 2 & 1 \\ 1 & -1 & 2 \end{vmatrix} \\=i\begin{vmatrix} 2 & 1\\ -1 & 2 \end{vmatrix} -j\begin{vmatrix} 1 & 1\\ 1 & 2 \end{vmatrix} + k\begin{vmatrix} 1 & 2\\ 1 & -1 \end{vmatrix}=5i-j-3k \\\text{The unit vector is} \\=\frac{5i-j-3k}{\sqrt{5^2+(-1)^2+(-3)^2}}=\frac{5i-j-3k}{\sqrt{35}}$$\text{The vector orthogonal to both (1, 2, 1) and (1, -1, 2) is the vector cross} \\\text{product between (1,2,1) and (1,-1,2)} \\\begin{vmatrix} i & j & k\\ 1 & 2 & 1 \\ 1 & -1 & 2 \end{vmatrix} \\=i\begin{vmatrix} 2 & 1\\ -1 & 2 \end{vmatrix} -j\begin{vmatrix} 1 & 1\\ 1 & 2 \end{vmatrix} + k\begin{vmatrix} 1 & 2\\ 1 & -1 \end{vmatrix}=5i-j-3k \\\text{The unit vector is} \\=\frac{5i-j-3k}{\sqrt{5^2+(-1)^2+(-3)^2}}=\frac{5i-j-3k}{\sqrt{35}}$

Question 2 isn't complete