Given:
a ⃗ = 4 i ^ + 3 j ^ \vec{a}=4\hat{i}+3\hat{j} a = 4 i ^ + 3 j ^
b ⃗ = 3 i ^ − 4 j ^ + 5 k ^ \vec{b}=3\hat{i}-4\hat{j}+5\hat{k} b = 3 i ^ − 4 j ^ + 5 k ^
So,
a ⃗ × b ⃗ = ∣ i ^ j ^ k ^ 4 3 0 3 − 4 5 ∣ \vec{a}\times \vec{b}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5\end{matrix}| a × b = ∣ i ^ 4 3 j ^ 3 − 4 k ^ 0 5 ∣
⇒ a ⃗ × b ⃗ = 15 i ^ − 20 j ^ − 25 k ^ \Rightarrow \vec{a}\times \vec{b}=15\hat{i}-20\hat{j}-25\hat{k} ⇒ a × b = 15 i ^ − 20 j ^ − 25 k ^
Let c ⃗ = x i ^ + y j ^ + z k ^ \vec{c}=x\hat{i}+y\hat{j}+z\hat{k} c = x i ^ + y j ^ + z k ^
Then,
c ⃗ ⋅ ( a ⃗ × b ⃗ ) + 25 = 0 \vec{c}\cdot (\vec{a}\times \vec{b})+25=0 c ⋅ ( a × b ) + 25 = 0
⇒ 15 x − 20 y − 25 z + 25 = 0 \Rightarrow 15x-20y-25z+25=0 ⇒ 15 x − 20 y − 25 z + 25 = 0
⇒ 3 x − 4 y − 5 z = − 5... ( 1 ) \Rightarrow 3x-4y-5z=-5...(1) ⇒ 3 x − 4 y − 5 z = − 5... ( 1 )
Also,
c ⃗ ⋅ ( i ^ + j ^ + k ^ ) = 4 \vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=4 c ⋅ ( i ^ + j ^ + k ^ ) = 4
⇒ x + y + z = 4.... ( 2 ) \Rightarrow x+y+z=4....(2) ⇒ x + y + z = 4.... ( 2 )
And projection of c ⃗ \vec{c} c on a ⃗ \vec{a} a is
c ⃗ ⋅ a ⃗ ∣ a ⃗ ∣ = 1 \frac{\vec{c}\cdot \vec{a}}{|\vec{a}|}=1 ∣ a ∣ c ⋅ a = 1
⇒ ( x i ^ + y j ^ + z k ^ ) ⋅ ( 4 i ^ + 3 j ^ ) 16 + 9 = 1 \Rightarrow \frac{(x\hat{i}+y\hat{j}+z\hat{k})\cdot (4\hat{i}+3\hat{j})}{\sqrt{16+9}}=1 ⇒ 16 + 9 ( x i ^ + y j ^ + z k ^ ) ⋅ ( 4 i ^ + 3 j ^ ) = 1
⇒ 4 x + 3 y = 5.... ( 3 ) \Rightarrow 4x+3y=5....(3) ⇒ 4 x + 3 y = 5.... ( 3 )
Solving (1),(2)&(3) , we get
x = 2 , y = − 1 , z = 3 x=2,y=-1,z=3 x = 2 , y = − 1 , z = 3
So,
c ⃗ = 2 i ^ − j ^ + 3 k ^ \vec{c}=2\hat{i}-\hat{j}+3\hat{k} c = 2 i ^ − j ^ + 3 k ^
Projection of c ⃗ \vec{c} c on b ⃗ \vec{b} b is
c ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ = ( 2 i ^ − j ^ + 3 k ^ ) ⋅ ( 3 i ^ − 4 j ^ + 5 k ^ ) 9 + 16 + 25 = 25 5 2 = 5 2 \frac{\vec{c}\cdot \vec{b}}{|\vec{b}|}=\frac{(2\hat{i}-\hat{j}+3\hat{k})\cdot (3\hat{i}-4\hat{j}+5\hat{k})}{\sqrt{9+16+25}}=\frac{25}{5\sqrt{2}}=\frac{5}{\sqrt{2}} ∣ b ∣ c ⋅ b = 9 + 16 + 25 ( 2 i ^ − j ^ + 3 k ^ ) ⋅ ( 3 i ^ − 4 j ^ + 5 k ^ ) = 5 2 25 = 2 5