Let a→=4i^+3j^,b→=3i^−4j^+5k^ . If c→ is a vector such that c→⋅(a→×b→)+25=0, c→⋅(i^+j^+k^)=4, and Projection of c→ on a→ is 1, then the projection of c→ on b→ equals

a¯×b¯=i^j^k^4303−45=15i^−20j^−25k^⇒|a¯×b¯|2=1250c¯=xa¯+yb¯+Z(a¯×b¯)c¯⋅a¯=5⇒5=x|a¯|2=25x∴x=15c⋅(a×b)=z|a¯×b¯|2=1250z=−25⇒z=−150c¯=45i^+35j^+yb¯+−150(15i^−20j^−25k^)c¯=45+3y−310i^+35−4y+z5j^+5y+1zk^c(i^+j^+k^)=4⇒12+3y+1−4y+5y+1z=4⇒4y=2⇒y=12∴c¯⋅b¯|b¯|=y|b¯|=52

Let a→=4i^+3j^,b→=3i^−4j^+5k^ . If c→ is a vector such that c→⋅(a→×b→)+25=0, c→⋅(i^+j^+k^)=4, and Projection of c→ on a→ is 1, then the projection of c→ on b→ equals

a¯×b¯=i^j^k^4303−45=15i^−20j^−25k^⇒|a¯×b¯|2=1250c¯=xa¯+yb¯+Z(a¯×b¯)c¯⋅a¯=5⇒5=x|a¯|2=25x∴x=15c⋅(a×b)=z|a¯×b¯|2=1250z=−25⇒z=−150c¯=45i^+35j^+yb¯+−150(15i^−20j^−25k^)c¯=45+3y−310i^+35−4y+z5j^+5y+1zk^c(i^+j^+k^)=4⇒12+3y+1−4y+5y+1z=4⇒4y=2⇒y=12∴c¯⋅b¯|b¯|=y|b¯|=52

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Let $\vec{a} = 4 \hat{i} + 3 \hat{j}, \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ . If $\vec{c}$ is a vector such that $\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$ , $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , and Projection of $\vec{c} on \vec{a} is 1$ , then the projection of $\vec{c} on \vec{b}$ equals

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$\frac{1}{\sqrt{2}}$

$\frac{1}{5}$

$\frac{3}{\sqrt{2}}$

$\frac{5}{\sqrt{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{a} \times \bar{b} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & - 4 & 5 \end{array}| = 15 \hat{i} - 20 \hat{j} - 25 \hat{k} \Rightarrow | \bar{a} \times \bar{b} |^{2} = 1250 \\ \bar{c} = x \bar{a} + y \bar{b} + Z (\bar{a} \times \bar{b}) \\ \bar{c} \cdot \bar{a} = 5 \Rightarrow 5 = x | \bar{a} |^{2} = 25 x ∴ x = \frac{1}{5} \\ c \cdot (a \times b) = z | \bar{a} \times \bar{b} |^{2} = 1250 z = - 25 \Rightarrow z = - \frac{1}{50} \\ \bar{c} = \frac{4}{5} \hat{i} + \frac{3}{5} \hat{j} + y \bar{b} + - \frac{1}{50} (15 \hat{i} - 20 \hat{j} - 25 \hat{k}) \\ \bar{c} = (\frac{4}{5} + 3 y - \frac{3}{10}) \hat{i} + (\frac{3}{5} - 4 y + \frac{z}{5}) \hat{j} + (5 y + \frac{1}{z}) \hat{k} \\ c (\hat{i} + \hat{j} + \hat{k}) = 4 \Rightarrow \frac{1}{2} + 3 y + 1 - 4 y + 5 y + \frac{1}{z} = 4 \Rightarrow 4 y = 2 \Rightarrow y = \frac{1}{2} \\ ∴ \frac{\bar{c} \cdot \bar{b}}{| \bar{b} |} = y | \bar{b} | = \frac{5}{\sqrt{2}} \end{array}$