Given vectors
a ⃗ = i ^ + 2 j ^ + k ^ = ( 1 , 2 , 1 ) , b ⃗ = 3 ( i ^ − j ^ + k ^ ) = ( 3 , − 3 , 3 ) \vec a=\hat i+2\hat j+\hat k=(1,2,1), \qquad \vec b=3(\hat i-\hat j+\hat k)=(3,-3,3) a = i ^ + 2 j ^ + k ^ = ( 1 , 2 , 1 ) , b = 3 ( i ^ − j ^ + k ^ ) = ( 3 , − 3 , 3 )
We are also given
a ⃗ × c ⃗ = b ⃗ , a ⃗ ⋅ c ⃗ = 3. \vec a\times \vec c=\vec b, \qquad \vec a\cdot \vec c=3. a × c = b , a ⋅ c = 3.
We need to find
a ⃗ ⋅ ( ( c ⃗ × b ⃗ ) − b ⃗ − c ⃗ ) . \vec a\cdot\big((\vec c\times \vec b)-\vec b-\vec c\big). a ⋅ ( ( c × b ) − b − c ) .
Break the required expression
Using distributivity of dot product,
a ⃗ ⋅ ( ( c ⃗ × b ⃗ ) − b ⃗ − c ⃗ ) = a ⃗ ⋅ ( c ⃗ × b ⃗ ) − a ⃗ ⋅ b ⃗ − a ⃗ ⋅ c ⃗ . \vec a\cdot\big((\vec c\times \vec b)-\vec b-\vec c\big)=\vec a\cdot(\vec c\times \vec b)-\vec a\cdot\vec b-\vec a\cdot\vec c. a ⋅ ( ( c × b ) − b − c ) = a ⋅ ( c × b ) − a ⋅ b − a ⋅ c .
So we compute these three terms one by one.
Compute a ⃗ ⋅ b ⃗ \vec a\cdot\vec b a ⋅ b
a ⃗ ⋅ b ⃗ = ( 1 , 2 , 1 ) ⋅ ( 3 , − 3 , 3 ) = 1 ⋅ 3 + 2 ⋅ ( − 3 ) + 1 ⋅ 3 = 3 − 6 + 3 = 0. \vec a\cdot\vec b=(1,2,1)\cdot(3,-3,3)=1\cdot 3+2\cdot(-3)+1\cdot 3=3-6+3=0. a ⋅ b = ( 1 , 2 , 1 ) ⋅ ( 3 , − 3 , 3 ) = 1 ⋅ 3 + 2 ⋅ ( − 3 ) + 1 ⋅ 3 = 3 − 6 + 3 = 0.
Hence,
a ⃗ ⋅ b ⃗ = 0. \vec a\cdot\vec b=0. a ⋅ b = 0.
Also given,
a ⃗ ⋅ c ⃗ = 3. \vec a\cdot\vec c=3. a ⋅ c = 3.
Compute a ⃗ ⋅ ( c ⃗ × b ⃗ ) \vec a\cdot(\vec c\times\vec b) a ⋅ ( c × b )
Use scalar triple product identity:
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = c ⃗ ⋅ ( b ⃗ × a ⃗ ) . \vec a\cdot(\vec c\times\vec b)=\vec c\cdot(\vec b\times\vec a). a ⋅ ( c × b ) = c ⋅ ( b × a ) .
Since b ⃗ = a ⃗ × c ⃗ \vec b=\vec a\times\vec c b = a × c , we can write
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = a ⃗ ⋅ ( c ⃗ × ( a ⃗ × c ⃗ ) ) . \vec a\cdot(\vec c\times\vec b)=\vec a\cdot\big(\vec c\times(\vec a\times\vec c)\big). a ⋅ ( c × b ) = a ⋅ ( c × ( a × c ) ) .
Now use the vector triple product identity:
x ⃗ × ( y ⃗ × z ⃗ ) = y ⃗ ( x ⃗ ⋅ z ⃗ ) − z ⃗ ( x ⃗ ⋅ y ⃗ ) . \vec x\times(\vec y\times\vec z)=\vec y(\vec x\cdot\vec z)-\vec z(\vec x\cdot\vec y). x × ( y × z ) = y ( x ⋅ z ) − z ( x ⋅ y ) .
With x ⃗ = c ⃗ , y ⃗ = a ⃗ , z ⃗ = c ⃗ \vec x=\vec c,\ \vec y=\vec a,\ \vec z=\vec c x = c , y = a , z = c ,
c ⃗ × ( a ⃗ × c ⃗ ) = a ⃗ ( c ⃗ ⋅ c ⃗ ) − c ⃗ ( c ⃗ ⋅ a ⃗ ) . \vec c\times(\vec a\times\vec c)=\vec a(\vec c\cdot\vec c)-\vec c(\vec c\cdot\vec a). c × ( a × c ) = a ( c ⋅ c ) − c ( c ⋅ a ) .
Therefore,
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = a ⃗ ⋅ [ a ⃗ ( c ⃗ ⋅ c ⃗ ) − c ⃗ ( c ⃗ ⋅ a ⃗ ) ] . \vec a\cdot(\vec c\times\vec b)=\vec a\cdot\left[\vec a(\vec c\cdot\vec c)-\vec c(\vec c\cdot\vec a)\right]. a ⋅ ( c × b ) = a ⋅ [ a ( c ⋅ c ) − c ( c ⋅ a ) ] .
So,
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = ( a ⃗ ⋅ a ⃗ ) ( c ⃗ ⋅ c ⃗ ) − ( a ⃗ ⋅ c ⃗ ) 2 . \vec a\cdot(\vec c\times\vec b)=(\vec a\cdot\vec a)(\vec c\cdot\vec c)-(\vec a\cdot\vec c)^2. a ⋅ ( c × b ) = ( a ⋅ a ) ( c ⋅ c ) − ( a ⋅ c ) 2 .
Now,
a ⃗ × c ⃗ = b ⃗ ⟹ ∣ a ⃗ × c ⃗ ∣ 2 = ∣ b ⃗ ∣ 2 . \vec a\times\vec c=\vec b \implies |\vec a\times\vec c|^2=|\vec b|^2. a × c = b ⟹ ∣ a × c ∣ 2 = ∣ b ∣ 2 .
But
∣ a ⃗ × c ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 ∣ c ⃗ ∣ 2 − ( a ⃗ ⋅ c ⃗ ) 2 . |\vec a\times\vec c|^2=|\vec a|^2|\vec c|^2-(\vec a\cdot\vec c)^2. ∣ a × c ∣ 2 = ∣ a ∣ 2 ∣ c ∣ 2 − ( a ⋅ c ) 2 .
Hence,
( a ⃗ ⋅ a ⃗ ) ( c ⃗ ⋅ c ⃗ ) − ( a ⃗ ⋅ c ⃗ ) 2 = ∣ b ⃗ ∣ 2 . (\vec a\cdot\vec a)(\vec c\cdot\vec c)-(\vec a\cdot\vec c)^2=|\vec b|^2. ( a ⋅ a ) ( c ⋅ c ) − ( a ⋅ c ) 2 = ∣ b ∣ 2 .
Thus,
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = ∣ b ⃗ ∣ 2 . \vec a\cdot(\vec c\times\vec b)=|\vec b|^2. a ⋅ ( c × b ) = ∣ b ∣ 2 .
Now calculate ∣ b ⃗ ∣ 2 |\vec b|^2 ∣ b ∣ 2 :
∣ b ⃗ ∣ 2 = 3 2 + ( − 3 ) 2 + 3 2 = 9 + 9 + 9 = 27. |\vec b|^2=3^2+(-3)^2+3^2=9+9+9=27. ∣ b ∣ 2 = 3 2 + ( − 3 ) 2 + 3 2 = 9 + 9 + 9 = 27.
So,
a ⃗ ⋅ ( c ⃗ × b ⃗ ) = 27. \vec a\cdot(\vec c\times\vec b)=27. a ⋅ ( c × b ) = 27.
Substitute into the required expression
a ⃗ ⋅ ( ( c ⃗ × b ⃗ ) − b ⃗ − c ⃗ ) = 27 − 0 − 3 = 24. \vec a\cdot\big((\vec c\times\vec b)-\vec b-\vec c\big)=27-0-3=24. a ⋅ ( ( c × b ) − b − c ) = 27 − 0 − 3 = 24.
Check options
The value is
24 \boxed{24} 24
So the correct option is C .