3.11 Vector equation of a line

3.11 Vector equation of a line Quiz

IB Maths: Applications and Interpretation HL: Quiz

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  1. 1. Which is a direction vector of r=(102)+λ(−246)\mathbf{r}=\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+\lambda\begin{pmatrix}-2 \\ 4 \\ 6\end{pmatrix}?

    • A. (102)\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}
    • B. (−123)\begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}
    • C. (−24−6)\begin{pmatrix}-2 \\ 4 \\ -6\end{pmatrix}
    • D. (2−46)\begin{pmatrix}2 \\ -4 \\ 6\end{pmatrix}

  2. 2. Which is a vector equation of the line through (3,−1,2)(3,-1,2) parallel to (14−2)\begin{pmatrix}1 \\ 4 \\ -2\end{pmatrix}?

    • A. r=(14−2)+λ(3−12)\mathbf{r}=\begin{pmatrix}1 \\ 4 \\ -2\end{pmatrix}+\lambda\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}
    • B. r=(3−12)+λ(−1−4−2)\mathbf{r}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}+\lambda\begin{pmatrix}-1 \\ -4 \\ -2\end{pmatrix}
    • C. r=(3−12)+λ(142)\mathbf{r}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}+\lambda\begin{pmatrix}1 \\ 4 \\ 2\end{pmatrix}
    • D. r=(3−12)+λ(14−2)\mathbf{r}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}+\lambda\begin{pmatrix}1 \\ 4 \\ -2\end{pmatrix}

  3. 3. A line has x=2+5λx=2+5\lambda, y=−3λy=-3\lambda, z=1+λz=1+\lambda. What are the coordinates of the point where λ=−1\lambda=-1?

    • A. (−3, 3, 0)(-3,\,3,\,0)
    • B. (7, −3, 2)(7,\,-3,\,2)
    • C. (−3, −3, 0)(-3,\,-3,\,0)
    • D. (−5, 3, 2)(-5,\,3,\,2)

  4. 4. Which point lies on r=(12)+λ(2−1)\mathbf{r}=\begin{pmatrix}1 \\ 2\end{pmatrix}+\lambda\begin{pmatrix}2 \\ -1\end{pmatrix}?

    • A. (5, 1)(5,\,1)
    • B. (3, 4)(3,\,4)
    • C. (7, −1)(7,\,-1)
    • D. (4, 0)(4,\,0)

  5. 5. A line has x=4−2λx=4-2\lambda, y=1+λy=1+\lambda, z=3λz=3\lambda. Which is its vector equation?

    • A. r=(410)+λ(213)\mathbf{r}=\begin{pmatrix}4 \\ 1 \\ 0\end{pmatrix}+\lambda\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}
    • B. r=(−213)+λ(410)\mathbf{r}=\begin{pmatrix}-2 \\ 1 \\ 3\end{pmatrix}+\lambda\begin{pmatrix}4 \\ 1 \\ 0\end{pmatrix}
    • C. r=(413)+λ(−210)\mathbf{r}=\begin{pmatrix}4 \\ 1 \\ 3\end{pmatrix}+\lambda\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}
    • D. r=(410)+λ(−213)\mathbf{r}=\begin{pmatrix}4 \\ 1 \\ 0\end{pmatrix}+\lambda\begin{pmatrix}-2 \\ 1 \\ 3\end{pmatrix}

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