Use the formula in Exercise 12.4.45 to find the distance from the point to the given line.
$ (0, 1, 3) ; x = 2t , y = 6 - 2t , z = 3 + t $
in the question they're asking do Use the formula in exercise 12 4 45 Question to find the distance from the point to the given line. The point is 013. And the equation of the line is x equal to two t. by equal to 6 -2 t. And the equally tripolis city. Therefore in order to find the distance of the point from the line first we have to find yeah the equation of this line in a regular form. So this uh means that the normal victor for the line. Mhm is equal to the coefficients of the tv level. That is too -2 and one while the point that passes through the line yeah is equal to zero minus Blessed six and 3. So this is the point and the victors through the line. Now according to the question we have to find the yeah distance. So in order to find the distance we have to use a formula that is given in the exercise. The formula is the is equal to a vector crossed be vector divided by a vector. So the scalar of the cross product in the enumerate. And the scalar of the a victor. Where a victor is the normal to the line. Yeah. Mhm. Uh huh. And v victor is mhm. So be vector is the mhm mm victor which is contained in the line which can be found out by using the formula the coordinates of the yeah yeah of the line. Mhm subtracted bye. The point given him the question therefore Herbie vector is equal to yeah. Yeah. Mhm. Yeah zero minus zero cuomo one minus six comma three minus three. That is yeah that is equal to zero comma minus five comma zero. Therefore this is the v victor. So after getting the values of A. N. D we can find. All right. The okay. Cross product of A. N. B. According to the formula. This is equal to the table I. J. Key and the values of A is 2 -2. 1. That is the direction vector of the line given in the question and be as though victor. That is yes derived from the two points of the line. And the point given the question that is 0 -50. And therefore it is equal to cap five minus Jacob zero plus kick up minus 10. So this is equal to 50 -10. Now in order to find the distance from the point 013 to the line given in the question. It is equal to scalar of the cross product of A and B divided by the scalar of a victor. So this is equal to the numerator. The scalar value of the cross product will be five square plus zero square plus 10 square divided by this killer value of the director. That is route under two square plus two squared plus one square. That is equal to route under 1 25 divided by route under nine. Now this is equal to five. Route five divided by three. Therefore the answer to this question is the distance mhm. Of the .013 to the line given in the question has a distance V is equal to five. Route 5 divided by three, and this is the required answer of the given question. Mhm.