12 त्रिविमीय ज्यामिति का परिचय
अभ्यास 3
Question:
Find the coordinates of the point which divides the line segment joining the points (−2,3,5) and (1,−4,6) in the ratio (i) 2:3 internally (ii) 2:3 externally.
Answer:
(i) Internally:
Let the coordinates of the point be (x, y, z).
Then, according to the given condition,
\frac{|\overrightarrow{PQ}|}{|\overrightarrow{PR}|} = \frac{2}{3}
Where PQ and PR are the vectors joining the points (−2,3,5) and (1,−4,6) to the point (x,y,z) respectively.
Therefore,
|\overrightarrow{PQ}| = 2|\overrightarrow{PR}|
⇒ (x + 2)^2 + (y - 3)^2 + (z - 5)^2 = 4(x - 1)^2 + (y + 4)^2 + (z - 6)^2
⇒ (x + 2)^2 + (y - 3)^2 + (z - 5)^2 = 4(x^2 - 2x + 1) + (y^2 + 8y + 16) + (z^2 - 12z + 36)
⇒ x^2 + 4x + 4 + y^2 - 6y + 9 + z^2 - 10z + 25 = 4x^2 - 8x + 4 + y^2 + 8y + 16 + z^2 - 12z + 36
⇒ -3x^2 - 4x + y^2 - 6y + z^2 - 10z + 9 = 0
⇒ x^2 + 4x + y^2 - 6y + z^2 - 10z = -9
⇒ x^2 + 4x + 4 + y^2 - 6y + 9 + z^2 - 10z + 25 = -9 + 25
⇒ x^2 + 4x + y^2 - 6y + z^2 - 10z = 16
⇒ x^2 + 4x + 16 + y^2 - 6y + 36 + z^2 - 10z = 16 + 36
⇒ x^2 + 4x + 16 + y^2 - 6y + 36 + z^2 - 10z = 52
Comparing the coefficients of x^2, y^2 and z^2, we get
x = -2, y = 1 and z = 4
Therefore, the coordinates of the point which divides the line segment joining the points (−2,3,5) and (1,−4,6) internally in the ratio 2:3 is (−2, 1, 4).
(ii) Externally:
Let the coordinates of the point be (x, y, z).
Then, according to the given condition,
\frac{|\overrightarrow{PQ}|}{|\overrightarrow{PR}|} = \frac{2}{3}
Where PQ and PR are the vectors joining the points (−2,3,5) and (1,−4,6) to the point (x,y,z) respectively.
Therefore,
|\overrightarrow{PQ}| = \frac{2}{3}|\overrightarrow{PR}|
⇒ (x + 2)^2 + (y - 3)^2 + (z - 5)^2 = \frac{4}{9}(x - 1)^2 + \frac{1}{9}(y + 4)^2 + \frac{1}{9}(z - 6)^2
⇒ (x + 2)^2 + (y - 3)^2 + (z - 5)^2 = \frac{4}{9}(x^2 - 2x + 1) + \frac{1}{9}(y^2 + 8y + 16) + \frac{1}{9}(z^2 - 12z + 36)
⇒ x^2 + 4x + 4 + y^2 - 6y + 9 + z^2 - 10z + 25 = \frac{4}{9}(x^2 - 2x + 1) + \frac{1}{9}(y^2 + 8y + 16) + \frac{1}{9}(z^2 - 12z + 36)
⇒ \frac{9}{4}x^2 + 6x + \frac{9}{4} + \frac{9}{4}y^2 - \frac{9}{2}y + \frac{9}{4}z^2 - \
Question:
Using section formula show that the points A (2,−3,4),B (−1,2,1) and C(0,1/3,2) are collinear.
Answer:
Step 1: Find the length of AB
AB = √((−1−2)2 + (2−(−3))2 + (1−4)2 )
= √(32 + 52 + (−3)2 )
= √(9 + 25 + 9)
= √43
Step 2: Find the length of BC
BC = √((0−(−1))2 + (1/3−2)2 + (2−1)2 )
= √(12 + (−5/3)2 + 12 )
= √(1 + 25/9 + 1)
= √(27/9)
Step 3: Find the length of AC
AC = √((2−0)2 + (−3−1/3)2 + (4−2)2 )
= √(22 + (−19/3)2 + 22 )
= √(4 + 361/9 + 4)
= √(369/9)
Step 4: Apply Section Formula
AB/AC = BC/BC
√43/√(369/9) = √(27/9)/√(27/9)
43/369 = 27/27
1 = 1
Hence, the points A (2,−3,4),B (−1,2,1) and C(0,1/3,2) are collinear.
Question:
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (−2,4,7) and (3,−5,8).
Answer:
Step 1: Calculate the midpoint of the line segment. Midpoint = ( ( -2 + 3 )/2 , ( 4 + -5 )/2 , ( 7 + 8 )/2 ) = ( (1)/2 , (-1)/2 , (15)/2 ) = ( 0.5 , -0.5 , 7.5 )
Step 2: Calculate the distance between the two points. Distance = √((3-(-2))2 + (-5-4)2 + (8-7)2) = √(52 + (-9)2 + 12) = √(25 + 81 + 1) = √107
Step 3: Calculate the ratio in which the YZ-plane divides the line segment. Ratio = Distance from midpoint to (−2,4,7) / Total Distance = √((-2-0.5)2 + (4-(-0.5))2 + (7-7.5)2) / √107 = √(2.52 + 4.52 + (-0.5)2) / √107 = √(6.25 + 20.25 + 0.25) / √107 = √(26.75) / √107 = 5.2 / 10.3 = 0.5048543689320389
Question:
Find the coordinates of the points which trisect the line segment joining the points P(4,2,−6) and Q(10,−16,6)
Answer:
-
Calculate the midpoint of the line segment joining the points P and Q by using the formula: Midpoint = (P + Q)/2
-
Substitute the given coordinates in the formula: Midpoint = ((4,2,-6) + (10,-16,6))/2
-
Simplify the equation to get the coordinates of the midpoint: Midpoint = (14,-7,0)/2
-
Divide each coordinate by 2 to get the coordinates of the midpoint: Midpoint = (7,-3.5,0)
Therefore, the coordinates of the points which trisect the line segment joining the points P(4,2,-6) and Q(10,-16,6) are (7,-3.5,0).
Question:
Given that P(3,2,−4),Q(5,4,−6) and R(9,8,−10) are collinear. Find the ratio in which Q divides PR.
Answer:
Step 1: Calculate the distance between points P and Q. Distance between P and Q = √((5-3)^2 + (4-2)^2 + (-6-(-4))^2) = √(4 + 4 + 4) = √12 = 2√3
Step 2: Calculate the distance between points P and R. Distance between P and R = √((9-3)^2 + (8-2)^2 + (-10-(-4))^2) = √(36 + 36 + 36) = √108 = 6√3
Step 3: Calculate the ratio in which Q divides PR. Ratio in which Q divides PR = Distance between P and Q / Distance between P and R = 2√3 / 6√3 = 1/3
जेईई अध्ययन सामग्री (गणित)
01 सेट
02 संबंध एवं फलन
03 त्रिकोणमितीय फलन
04 गणितीय आगमन का सिद्धांत
05 सम्मिश्र संख्याएँ और द्विघात समीकरण
06 रैखिक असमानताएँ
07 क्रमचय और संचय
08 द्विपद प्रमेय
09 अनुक्रम और श्रृंखला
10 सीधी रेखाओं का अभ्यास
10 सीधी रेखाएँ विविध
11 शांकव खंड
12 त्रिविमीय ज्यामिति का परिचय
13 सीमाएं और डेरिवेटिव
14 गणितीय तर्क
15 सांख्यिकी
16 प्रायिकता