The point $A$ has coordinates $(1, -1, 2)$ and the line $l$ has equation \(\mathbf{r} = 2\math

Question

The point $A$ has coordinates $(1, -1, 2)$ and the line $l$ has equation $\mathbf{r} = 2\mathbf{i} - 3\mathbf{j} + \lambda(\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})$. The point $A$ and the line $l$ lies in the plane $p_1$. Find the position vector of the foot of perpendicular from $A$ to the line $l$. [3] Show that the equation of $p_1$ is $2x + y = 1$.[2] The plane $p_2$ with equation $8x + ay + z = 4$ meets the plane $p_1$ at the line $L$ and the point $B(0, 1, 0)$ lies on $L$. Show that $L$ is parallel to $\mathbf{i} - 2\mathbf{j} + (2a - 8)\mathbf{k}$.[1] Show that $a = 4$.[1] Find the angle between $l$ and $p_2$.[2] Another point $C$ lies in the plane $p_1$ such that $BC$ is perpendicular to the line $L$. Show that $\overrightarrow{BC}$ is parallel to $\mathbf{k}$. Given that the distance of $C$ from $p_2$ is 5 units, find the possible coordinates of $C$.[4]

Accepted Answer

(i) $\dfrac16\begin{pmatrix}13\-20\5\end{pmatrix}$ (ii) $2x+y=1$ (iii) $\mathbf i-2\mathbf j+(2a-8)\mathbf k$ (iv) $a=4$ (v) $5.82^\circ$ (vi) $C=(0,1,45)$ or $C=(0,1,-45)$

Vectors

Vectors

Video Solution:

Solution:

Solution locked