The unit vector $\mathbf{d}$ makes angles of ${{60}^{\circ }}$ with both the $x$- and $y$-axes, and

Question

The unit vector $\mathbf{d}$ makes angles of ${{60}^{\circ }}$ with both the $x$- and $y$-axes, and $	heta $ with the $z$-axis, where ${{0}^{\circ }}\le 	heta \le {{90}^{\circ }}$. Show that $\mathbf{d}$ is parallel to $\mathbf{i}+\mathbf{j}+\sqrt{2}\mathbf{k}$. [3] The line $m$ is parallel to $\mathbf{d}$and passes through the point with coordinates $\left( 2,-1,0 ight)$. Find the coordinates of the point on $m$ that is closest to the point with coordinates $\left( 3,2,0 ight)$. [3] The plane ${{p}_{1}}$ has equation $\mathbf{r}\cdot \left( \mathbf{i}+2\mathbf{j}+2\mathbf{k} ight)=5$, and the line $l$ has equation $\frac{x-a}{2}=\frac{y-1}{b}=-\frac{z}{2}$, where $a$ and $b$ are constants. Given that $l$ lies on ${{p}_{1}}$, show that $b=1$ and find the value of $a$. [2] The plane ${{p}_{2}}$ contains $l$ and is perpendicular to ${{p}_{1}}$. Find the equation of ${{p}_{2}}$ in the form $\mathbf{r}\cdot \mathbf{n}=c$, where $c$ is a constant to be determined. [3] The variable point $P\left( x,y,z ight)$ is equidistant from ${{p}_{1}}$ and ${{p}_{2}}$. Find the cartesian equation(s) of the locus of $P$. [3]

Accepted Answer

(a)(i) $\mathbf d\parallel(1,1,\sqrt2)$ (ii) $F=(3,0,\sqrt2)$ (b) $a=3$, $b=1$ (i) $p_2:\mathbf r\cdot\begin{pmatrix}2\-2\1\end{pmatrix}=4$ (ii) $x-4y-z=-1$ or $x+z=3$

2017 ACJC P2 Q4

2017 ACJC P2 Q4

Video Solution:

Solution:

Solution locked