1.
If two vectors are perpendicular then A → . B → = 0 \overrightarrow{A}.\overrightarrow{B}=0 A . B = 0
[ − k i ^ + ( i + k ) j ^ + 9 k + k 2 ) k ^ ] . [ ( 1 + k ) i ^ + ( k + k 2 ) j ^ − k k ^ = 0 [-k\widehat{i}+(i+k)\widehat{j}+9k+k^2)\widehat{k}].[(1+k)\widehat{i}+(k+k^2)\widehat{j}-k\widehat{k}=0 [ − k i + ( i + k ) j + 9 k + k 2 ) k ] . [( 1 + k ) i + ( k + k 2 ) j − k k = 0
− k ( 1 + k ) + ( 1 + k ) ( k + k 2 ) + ( k + k 2 ) ( − k ) = 0 -k(1+k)+(1+k)(k+k^2)+(k+k^2)(-k)=0 − k ( 1 + k ) + ( 1 + k ) ( k + k 2 ) + ( k + k 2 ) ( − k ) = 0
− k − k 2 + k + k 2 + k 2 + k 3 − k 2 − k 3 = 0 0 = 0 -k-k^2+k+k^2+k^2+k^3-k^2-k^3=0\\0=0 − k − k 2 + k + k 2 + k 2 + k 3 − k 2 − k 3 = 0 0 = 0
option 1 is correct
2.
∣ A → ∣ = ( − k ) 2 + ( 1 + k ) 2 + ( k + k 2 ) |\overrightarrow{A}|=\sqrt{(-k)^2+(1+k)^2+(k+k^2)} ∣ A ∣ = ( − k ) 2 + ( 1 + k ) 2 + ( k + k 2 )
= k 2 + 1 + k 2 + 2 k + k 2 + k 4 + 2 k 3 \\=\sqrt{k^2+1+k^2+2k+k^2+k^4+2k^3} = k 2 + 1 + k 2 + 2 k + k 2 + k 4 + 2 k 3
= k 4 + 2 k 3 + 3 k 2 + 2 k + 1 =\sqrt{k^4+2k^3+3k^2+2k+1} = k 4 + 2 k 3 + 3 k 2 + 2 k + 1
∣ B → ∣ = ( 1 + k ) 2 + ( k + k 2 ) + ( − k ) 2 \\|\overrightarrow{B}|=\sqrt{(1+k)^2+(k+k^2)+(-k)^2} ∣ B ∣ = ( 1 + k ) 2 + ( k + k 2 ) + ( − k ) 2
= 1 + k 2 + 2 k + k 2 + k 4 + 2 k 3 + k 2 \\=\sqrt{1+k^2+2k+k^2+k^4+2k^3+k^2} = 1 + k 2 + 2 k + k 2 + k 4 + 2 k 3 + k 2
k 4 + 2 k 3 + 3 k 2 + 2 k + 1 \\\sqrt{k^4+2k^3+3k^2+2k+1} k 4 + 2 k 3 + 3 k 2 + 2 k + 1
∣ A → ∣ = ∣ B → ∣ \\|\overrightarrow{A}|=|\overrightarrow{B}| ∣ A ∣ = ∣ B ∣
option 2 is correct
3.
if A → & B → \overrightarrow{A} \& \overrightarrow{B} A & B is a unit vector
∣ A → ∣ = 1 , ∣ B → ∣ = 1 |\overrightarrow{A}|=1,\space |\overrightarrow{B}|=1 ∣ A ∣ = 1 , ∣ B ∣ = 1
k 4 + 2 k 3 + 3 k 2 + 2 k + 1 = 1 k ( k 3 + 2 k 2 + 3 k + 2 ) = 0................ ( 1 ) f o r k = 1 k^4+2k^3+3k^2+2k+1=1\\k(k^3+2k^2+3k+2)=0................(1)\\for\space k=1 k 4 + 2 k 3 + 3 k 2 + 2 k + 1 = 1 k ( k 3 + 2 k 2 + 3 k + 2 ) = 0................ ( 1 ) f or k = 1
equation (1) not satisfied so option (3) is not correct
4.
∣ A → ∣ = 1 |\overrightarrow{A}|=1 ∣ A ∣ = 1
by equation (1)
k ( ( k 3 + 2 k 2 + 3 k + 2 ) = 0 k = 0 k 3 + 2 k 2 + 3 k + 2 = 0 p u t k = − 1 − 1 + 2 − 3 + 2 = 0 0 = 0 k((k^3+2k^2+3k+2)=0\\k=0\\k^3+2k^2+3k+2=0\\put\space k=-1\\-1+2-3+2=0\\0=0 k (( k 3 + 2 k 2 + 3 k + 2 ) = 0 k = 0 k 3 + 2 k 2 + 3 k + 2 = 0 p u t k = − 1 − 1 + 2 − 3 + 2 = 0 0 = 0
option 4 is correct
5.
vector is parallel if
− k = d ( 1 + k ) & 1 + k = d ( k ( 1 + k ) − k 1 + k = d , 1 k = d & k ( 1 + k ) = d ( − k ) , d = ( − 1 + k ) -k=d(1+k)\space \&\space 1+k=d(k(1+k)\\\frac{-k}{1+k}=d,\space \frac{1}{k}=d\space \&\space k(1+k)=d(-k),\space d=(-1+k) − k = d ( 1 + k ) & 1 + k = d ( k ( 1 + k ) 1 + k − k = d , k 1 = d & k ( 1 + k ) = d ( − k ) , d = ( − 1 + k )
Therefore the vectors are not parallel
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