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Question
যদি
A
→
=
6
i
^
-
3
j
^
-
3
k
^
এবং
B
→
=
2
i
^
+
2
j
^
+
k
^
হয়, তবে
A
→
ও
B
→
মধ্যবর্তী কোণ কোনটি?1
69.98
°
70.98
°
71.98
°
72.98
°
ANSWER : 1
Descrption
<p style="margin-left:0px;">→A = 6ˆi - 3ˆj - 3ˆk →B = 2ˆi + 2ˆj + ˆk</p><p style="margin-left:0px;">→A ও →B এর মধ্যবর্তী কোণ θ হলে,</p><p style="margin-left:0px;">cos θ = (→A ⋅ →B)/(||→A|| ||→B||)</p><p style="margin-left:0px;">→A ⋅ →B = (6ˆi - 3ˆj - 3ˆk) ⋅ (2ˆi + 2ˆj + ˆk)</p><p style="margin-left:0px;">= 12 - 6 - 3</p><p style="margin-left:0px;">= 3</p><p style="margin-left:0px;">||→A|| = ||6ˆi - 3ˆj - 3ˆk||</p><p style="margin-left:0px;">= √(6^2 + (-3)^2 + (-3)^2)</p><p style="margin-left:0px;">= √45</p><p style="margin-left:0px;">= 3√5</p><p style="margin-left:0px;">||→B|| = ||2ˆi + 2ˆj + ˆk||</p><p style="margin-left:0px;">= √(2^2 + 2^2 + 1^2)</p><p style="margin-left:0px;">= √9</p><p style="margin-left:0px;">= 3</p><p style="margin-left:0px;">cos θ = 3/(3√5 * 3)</p><p style="margin-left:0px;">cos θ = 1/√5</p><p style="margin-left:0px;">θ = arccos(1/√5)</p><p style="margin-left:0px;">θ ≈ 69.98°</p><p style="margin-left:0px;"><strong>উত্তরঃ 69.98°</strong></p>
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