Please answer this question ASAP
Answers
i)
ii)
First let us calculate ,
- | A | , | B |
- A.B
a] |A| =
b] | B | =
c] A.B = (2×1) + (-1×2) + (-3×-1)
= 2 + (-2) + 3 = 3
Now ,
iii) First let us find the values of ,
a]
b]
Now ,
Answer:
\huge\red{\bold{\underline{\underline{Given:-}}}}
Given:−
\large\bold{\overrightarrow{A}\:=\:2\hat{i}-\hat{j}-3\hat{k}}
A
=2
i
^
−
j
^
−3
k
^
\large\bold{\overrightarrow{B}\:=\:\hat{i}+2\hat{j}-\hat{k}}
B
=
i
^
+2
j
^
−
k
^
\huge\blue{\bold{\underline{\underline{To\:Find:-}}}}
ToFind:−
\large\bold{\overrightarrow{A}\times\:\overrightarrow{B}}
A
×
B
\large\bold{|\overrightarrow{A}+\overrightarrow{B}|}∣
A
+
B
∣
\large\bold{(\overrightarrow{A}+\overrightarrow{B})\times\:({\overrightarrow{A}-\overrightarrow{B})}}(
A
+
B
)×(
A
−
B
)
\huge\green{\bold{\underline{\underline{Solution:-}}}}
Solution:−
i) \large\bold{\overrightarrow{A}\times\:\overrightarrow{B}\:=\:(2\hat{i}-\hat{j}-3\hat{k})\times\:(\hat{i}+2\hat{j}-\hat{k})}
A
×
B
=(2
i
^
−
j
^
−3
k
^
)×(
i
^
+2
j
^
−
k
^
)
\begin{gathered}\large\bold{=\:\begin{vmatrix} \sf \hat{i}& \sf\hat{-j}& \sf \hat{k} \\ \sf 2& \sf -1& \sf -3 \\ \sf 1& \sf 2& \sf -1 \end{vmatrix}}\end{gathered}
=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
2
1
−j
^
−1
2
k
^
−3
−1
∣
∣
∣
∣
∣
∣
∣
∣
\large\bold{=\:[(1)-(-6)]\hat{i}-[(-2)-(-3)]\hat{j}+[4-(-1)]\hat{k}}=[(1)−(−6)]
i
^
−[(−2)−(−3)]
j
^
+[4−(−1)]
k
^
\large\bold{=\:[7\hat{i}-\hat{j}+5\hat{k}]}=[7
i
^
−
j
^
+5
k
^
]
ii) \large\bold{|\overrightarrow{A}+\overrightarrow{B}|\:=\:\sqrt{|A|^2+|B|^2+2A.B}}∣
A
+
B
∣=
∣A∣
2
+∣B∣
2
+2A.B
First let us calculate ,
| A | , | B |
A.B
a] |A| = \large\bold{\sqrt{(2)^2+(-1)^2+(-3)^2}}
(2)
2
+(−1)
2
+(−3)
2
\large\bold{=\:\sqrt{14}}=
14
b] | B | = \large\bold{\sqrt{(1)^2+(-1)^2+(2)^2}}
(1)
2
+(−1)
2
+(2)
2
\large\bold{=\:\sqrt{6}}=
6
c] A.B = (2×1) + (-1×2) + (-3×-1)
= 2 + (-2) + 3 = 3
Now ,
\large\bold{|\overrightarrow{A}+\overrightarrow{B}|\:=\:\sqrt{(\sqrt{14})^2+(\sqrt{6})^2+2(3)}}∣
A
+
B
∣=
(
14
)
2
+(
6
)
2
+2(3)
\large\bold{=\:\sqrt{26}}=
26
iii) First let us find the values of ,
\large\bold{(\overrightarrow{A}+\overrightarrow{B})}(
A
+
B
)
\large\bold{(\overrightarrow{A}-\overrightarrow{B})}(
A
−
B
)
a] \large\bold{(\overrightarrow{A}+\overrightarrow{B})\:=\:(2\hat{i}-\hat{j}-3\hat{k})+(\hat{i}+2\hat{j}-\hat{k})}(
A
+
B
)=(2
i
^
−
j
^
−3
k
^
)+(
i
^
+2
j
^
−
k
^
)
\large\bold{=\:3\hat{i}+\hat{j}-4\hat{k}}=3
i
^
+
j
^
−4
k
^
b] \large\bold{(\overrightarrow{A}-\overrightarrow{B})\:=\:(2\hat{i}-\hat{j}-3\hat{k})-(\hat{i}+2\hat{j}-\hat{k})}(
A
−
B
)=(2
i
^
−
j
^
−3
k
^
)−(
i
^
+2
j
^
−
k
^
)
\large\bold{=\:\hat{i}-3\hat{j}-2\hat{k}}=
i
^
−3
j
^
−2
k
^
Now ,
\large\bold{(\overrightarrow{A}+\overrightarrow{B})\times\:(\overrightarrow{A}-\overrightarrow{B})\:=\:{(3\hat{i}+\hat{j}-4\hat{k})\times\:(\hat{i}-3\hat{j}-2\hat{k})}}(
A
+
B
)×(
A
−
B
)=(3
i
^
+
j
^
−4
k
^
)×(
i
^
−3
j
^
−2
k
^
)
\begin{gathered}\large\bold{=\:\begin{vmatrix} \sf \hat{i}& \sf\hat{-j}& \sf \hat{k} \\ \sf 3& \sf 1& \sf -4 \\ \sf 1& \sf -3& \sf -2 \end{vmatrix}}\end{gathered}
=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
3
1
−j
^
1
−3
k
^
−4
−2
∣
∣
∣
∣
∣
∣
∣
∣
\large\bold{=\:[(-2)-(12)]\hat{i}-[-6-(-4)]\hat{j}+[-9-(1)]\hat{k}}=[(−2)−(12)]
i
^
−[−6−(−4)]
j
^
+[−9−(1)]
k
^
\large\bold{=\:[-14\hat{i}+2\hat{j}-10\hat{k}]}=[−14
i
^
+2
j
^
−10
k
^
]
\large\bold{\therefore{\overrightarrow{A}\times\:\overrightarrow{B}\:=\:7\hat{i}-\hat{j}+5\hat{k}}}∴
A
×
B
=7
i
^
−
j
^
+5
k
^
\large\bold{|\overrightarrow{A}+\overrightarrow{B}|\:=\:\sqrt{26}}∣
A
+
B
∣=
26
\large\bold{(\overrightarrow{A}+\overrightarrow{B})\times\:(\overrightarrow{A}-\overrightarrow{B})\:=\:-14\hat{i}+2\hat{j}-10\hat{k}}(
A
+
B
)×(
A
−
B
)=−14
i
^
+2
j
^
−10
k
^