Question

4. [3/8 Points) DETAILS PREVIOUS ANSWERS SCALCCC4 9.5.013. Consider the line that passes through the point and is parallel to the given vector. (4, -3,6) < 1, 2, -3> (a) Find symmetric equations for the line. 2-6 -(x-4)= 2 3 y +3 (b) Find the points in which the line intersects the coordinate planes. (5 X IX 0) (0,9 X -6 (4 x 0, 4 X) 1

Answer 1

2-4 y +3 2 4. b) since the Symmatric equations for the line is g. zo Now we have to find the points in which the line intersects the coordinate planes. for the intersection with the rey-plane , set z=o. g Y+3 x - 4 0-6 --3 2 2. = 7+3 => 2-4 2 x-4 =-2 thaas s. 2-4 x- 个介 4-9. 2. anda y +3 2 = 2 => 7 +3 4. 4-3 = 1 - y o the point of intersection with the xy-plane 18 (29190) for the intersection with the az-plane, y=0 c. *=1 26 thus, 2-4 = 32 set -0. = 0 +3 2 -3. - 2 =>X-4 11 -2+4 Lolor 2 => and, 2 - 6 3 介 2-6 - 2 9 2 = -9 +6 =7 Z ola 3 2. 2

is ( 3 / g o the point of intersection with the xz plane 90,) for the intersection with yz-plane i set x = 0, Zb ó. 0-4 1 Y+3 2 y +3 =4 2 > y + 3 = 8 os 8-3 = 5: Z-6 = 4 z-6 11 -12 => z co => Z -12 +6. -6. the point of intersection with the y z-plane 0959-6). oo the points in which the line interseets coordinate planes a O,0), the 2 9 . 5 -6) o 2 9. 9 0 Mlog ofd 3. 2 9 9 2

** (6) since the parametrie equations for the line is, = 5+t: 3-t 6 +26: z = عه in for the az-plane, set y zo =) t o 3-t = 3. 2. oo by substitute we get, 24 = 5+3 8 9 Z = 6 H3)2 = 12 - 6 +6 ... the point of intersection with az-plane is (8,0, 12) set z =0 for the xy-plane , 6+2+ = 0 =) 27 oc -6. t -6s-3 2 is by substitute we gets a = 5-3 = 2 os xy-plane set x=0 9 io 5 ++ =0 3-(-3) = 6. with do the point of intersection (29690). for the yz - plane 5) t = -5 gets. = 3-6-5) = 3+5 = 8 6+2(-5) = 6-10 = -4 %o the point of intersection with yz-plane is (0989-4). oe oo by substitute y z =

so 9 the points in which the line intersects the coordinate planes. ( 0.9 8 |-41) ( [৪] 21), ( 2 , 61,0) 09 12 n 2 6 9

We know that a line g ou is, 9 we have to find the vector equation for the line segment from P to 0.70++1) the vector equation of a segment joining the points with position vectors ro and (1-t) 7. + tout det, ño = < 2, -3,4% and a = 55,593) ño It) (1-t) < 29-394) + t <5,593) - (2-2t+57 , -3+3++5.594-4€+3+) ñ 1) 20 o = 12 +34 9-3+8+,4-t> oo r(t) 2+37] î + | -3 +84 +14-47 -