09 ଭେକ୍ଟର୍
ବିବିଧ ସମାଧାନ
Question:
What pulled the Princess out of her gloom?
Answer:
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First, it is important to understand why the Princess was feeling gloomy.
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Once the reason for the Princess’ gloom is identified, it can be determined what could have pulled her out of her gloom.
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Possible solutions could include activities such as spending time with friends, engaging in hobbies, or going on an adventure.
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Ultimately, the exact solution to the Princess’ gloom will depend on the individual and the specific circumstances.
Question:
Write down a unit vector in XY- plane, making an angle of 30∘ with the positive direction of x−axis .
Answer:
A unit vector in XY-plane making an angle of 30∘ with the positive direction of x−axis can be written as:
u = (cos 30°, sin 30°) = (√3/2, 1/2)
Question:
Find the scalar components and magnitude of the vector joining the points P(x1,y1,z1) and Q(x2,y2,z2).
Answer:
Step 1: Calculate the difference between the x-coordinates of the two points: x2 - x1.
Step 2: Calculate the difference between the y-coordinates of the two points: y2 - y1.
Step 3: Calculate the difference between the z-coordinates of the two points: z2 - z1.
Step 4: Calculate the scalar components of the vector by taking the square root of the sum of the squares of the differences calculated in Steps 1-3:
√(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
Step 5: Calculate the magnitude of the vector by taking the square root of the sum of the squares of the scalar components:
√(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
Question:
Find the value of x for which x(i^+j^+k^) is a unit vector .
Answer:
Step 1: A unit vector is a vector with a magnitude of 1. Therefore, we need to find the magnitude of x(i^+j^+k^).
Step 2: The magnitude of a vector can be found using the formula: |v| = √(v₁² + v₂² + v₃²).
Step 3: Substitute i^+j^+k^ into the formula to get: |x(i^+j^+k^)| = √(x² + x² + x²) = √(3x²).
Step 4: Set the magnitude equal to 1 and solve for x.
|x(i^+j^+k^)| = 1
√(3x²) = 1
3x² = 1
x² = 1/3
x = √(1/3)
Question:
A girl walks 4 km towards west, then she walk 3 km in a direction 30o east of north and stops. Determine the girl’s displacement from her initial point of departure.
Answer:
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The girl’s initial point of departure is the origin (0,0).
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She walks 4 km towards west, so her new coordinates are (-4,0).
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She then walks 3 km in a direction 30o east of north. To find the new coordinates, we use the trigonometric ratio of sine and cosine.
sin 30o = 3/r cos 30o = 4/r
where r is the magnitude (or distance) of the girl’s displacement.
Therefore, r = 4√3
Using the trigonometric ratios, we can find the new x- and y-coordinates of the girl’s position.
x = -4 + 4√3 cos 30o = -4 + 4√3 (4/r) = -4 + 16/√3
y = 0 + 4√3 sin 30o = 0 + 4√3 (3/r) = 12/√3
- The girl’s displacement from her initial point of departure is the magnitude of the vector joining the two points, which is given by the Pythagorean theorem.
displacement = √((x2 - x1)2 + (y2 - y1)2)
= √((-4 + 16/√3)2 + (12/√3)2)
= √(256/3 + 144/3)
= √(400/3)
= 20√3 units
Question:
If a=b+c, then is it true that ∣a∣=∣b∣+∣c∣? Justify your answer.
Answer:
No, it is not true that |a|=|b|+|c|.
To justify this answer, we can look at an example. Let a=2, b=-1, and c=1.
Then a=b+c, or a=2.
However, |a|=|2|=2, while |b|+|c|=|-1|+|1|=2+1=3.
Since |a| does not equal |b|+|c|, it is not true that |a|=|b|+|c|.
Question:
If a=i^+j^+k^,b=2i^−j^+3k^ and c=i^−2j^+k^, find a unit vector parallel to the vector 2a−b+3c.
Answer:
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Expand the vectors: a = i + j + k b = 2i - j + 3k c = i - 2j + k
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Substitute the vectors into the equation: 2a - b + 3c = 2(i + j + k) - (2i - j + 3k) + 3(i - 2j + k)
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Simplify the equation: 2a - b + 3c = 2i + 2j + 2k - 2i + j - 3k + 3i - 6j + 3k
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Combine like terms: 2a - b + 3c = 4i - 5j + 6k
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Divide the equation by its magnitude: 2a - b + 3c = (4i - 5j + 6k) / √(42 + 52 + 62)
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Simplify the equation: 2a - b + 3c = (4/√77)i - (5/√77)j + (6/√77)k
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The unit vector parallel to the vector 2a - b + 3c is: (4/√77)i - (5/√77)j + (6/√77)k
JEE ଅଧ୍ୟୟନ ସାମଗ୍ରୀ (ଗଣିତ)
01 ସମ୍ପର୍କ ଏବଂ କାର୍ଯ୍ୟ
02 ଓଲଟା ଟ୍ରାଇଗୋନେଟ୍ରିକ୍ କାର୍ଯ୍ୟଗୁଡ଼ିକ
03 ମ୍ୟାଟ୍ରିକ୍ସ
04 ନିର୍ଣ୍ଣୟକାରୀ
05 ନିରନ୍ତରତା ଏବଂ ଭିନ୍ନତା
- ବ୍ୟାୟାମ 01
- ବ୍ୟାୟାମ 02
- ବ୍ୟାୟାମ 03
- ବ୍ୟାୟାମ 04
- ବ୍ୟାୟାମ 05
- ବ୍ୟାୟାମ 06
- ବ୍ୟାୟାମ 07
- ବ୍ୟାୟାମ 08
- ବିବିଧ ବ୍ୟାୟାମ
06 ଡେରିଭେଟିକ୍ସର ପ୍ରୟୋଗ
07 ଇଣ୍ଟିଗ୍ରାଲ୍
08 ଇଣ୍ଟିଗ୍ରାଲ୍ସର ପ୍ରୟୋଗ
09 ଭେକ୍ଟର୍
10 ତିନୋଟି ଡାଇମେନ୍ସନାଲ୍ ଜ୍ୟାମିତି
11 ରେଖା ପ୍ରୋଗ୍ରାମିଂ
12 ସମ୍ଭାବନା