
Find the distance of the point P $\left( {3,4,4} \right)$ from the point, where the line joining the point A $\left( {3, - 4, - 4} \right)$ and B $\left( {2, - 3,1} \right)$$\left( {2, - 3,1} \right)$ intersects the plane \[2x + y + z = 7\]
Answer
508.8k+ views
Hint: Since the points are given you can easily find the equation of line from the known points using proper formula. Then find the coordinates of the point which intersects the found line and the given equation of the plane. Once you find the point, you can find the distance of both points which is found by the distance formula. You can find the equation of the line passing through two points $\left( {{a_1},{b_1},{c_1}} \right)$ and $\left( {{a_2},{b_2},{c_2}} \right)$ by \[\dfrac{{x - {a_1}}}{{{a_2} - {a_1}}} = \dfrac{{y - {b_1}}}{{{b_2} - {b_1}}} = \dfrac{{z - {c_1}}}{{{c_2} - {c_1}}}\]
Complete step by step answer:
Given: Two points of line are given. They are A $\left( {3, - 4, - 4} \right)$) and B $\left( {2, - 3,1} \right)$. Equation of the plane is given to be \[2x + y + z = 7\]
From the formula given in hint, we find the equations as
\[\dfrac{{x - 3}}{{2 - 3}} = \dfrac{{y + 4}}{{ - 3 + 4}} = \dfrac{{z + 5}}{{1 + 5}}\]
Let, \[\dfrac{{x - 3}}{{2 - 3}} = \dfrac{{y + 4}}{{ - 3 + 4}} = \dfrac{{z + 5}}{{1 + 5}} = k\]
\[ \Rightarrow x = - k + 3,y = k - 4,z = 6k - 5\]
The coordinates of the point of intersection of given line and plane is \[( - k + 3,k - 4,6k - 5)\] - (i)
For this point to lie on the plane \[2x + y + z = 7\] , the point has to satisfy the equation of the plane. From the equation given, the value of k is found to be 2.
Putting k=2 in equation (i),
Coordinates of the point of intersection of the given line and plane is (1, -2, 7).
Therefore, we need to find the distance between $\left( { - 1,2,7} \right)$ and $\left( {3,4,4} \right)$.
By using distance formula, we find
Distance = \[\sqrt {{{(3 - 1)}^2} + {{(4 - 2)}^2} + {{(4 - 7)}^2}} = 7\] units.
Hence, the distance of P \[(3,4,4)\] from the points given is 7 units.
Note:
To solve this question, the basics of line and planes are very much needed. You will find a lot of questions in the NCERT book which are great for concept building. The knowledge of formulas is very much needed to solve these questions else you will get stuck in the question. Also visualise in rough what the question is asking about so that you can approach the question in the right manner.
Complete step by step answer:
Given: Two points of line are given. They are A $\left( {3, - 4, - 4} \right)$) and B $\left( {2, - 3,1} \right)$. Equation of the plane is given to be \[2x + y + z = 7\]
From the formula given in hint, we find the equations as
\[\dfrac{{x - 3}}{{2 - 3}} = \dfrac{{y + 4}}{{ - 3 + 4}} = \dfrac{{z + 5}}{{1 + 5}}\]
Let, \[\dfrac{{x - 3}}{{2 - 3}} = \dfrac{{y + 4}}{{ - 3 + 4}} = \dfrac{{z + 5}}{{1 + 5}} = k\]
\[ \Rightarrow x = - k + 3,y = k - 4,z = 6k - 5\]
The coordinates of the point of intersection of given line and plane is \[( - k + 3,k - 4,6k - 5)\] - (i)
For this point to lie on the plane \[2x + y + z = 7\] , the point has to satisfy the equation of the plane. From the equation given, the value of k is found to be 2.
Putting k=2 in equation (i),
Coordinates of the point of intersection of the given line and plane is (1, -2, 7).
Therefore, we need to find the distance between $\left( { - 1,2,7} \right)$ and $\left( {3,4,4} \right)$.
By using distance formula, we find
Distance = \[\sqrt {{{(3 - 1)}^2} + {{(4 - 2)}^2} + {{(4 - 7)}^2}} = 7\] units.
Hence, the distance of P \[(3,4,4)\] from the points given is 7 units.
Note:
To solve this question, the basics of line and planes are very much needed. You will find a lot of questions in the NCERT book which are great for concept building. The knowledge of formulas is very much needed to solve these questions else you will get stuck in the question. Also visualise in rough what the question is asking about so that you can approach the question in the right manner.
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