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\(A B C\) is a right-angled triangle in which \(\max \{A B, B C, A C\}=B C\). If the position vectors of \(B\) and \(C\) are respectively \(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}\), then
\[
\mathbf{A B} \cdot \mathbf{A C}+\mathbf{B A} \cdot \mathbf{B C}+\mathbf{C A} \cdot \mathbf{C B}=
\]
Options:
\[
\mathbf{A B} \cdot \mathbf{A C}+\mathbf{B A} \cdot \mathbf{B C}+\mathbf{C A} \cdot \mathbf{C B}=
\]
Solution:
1271 Upvotes
Verified Answer
The correct answer is:
29
Given
\(\begin{aligned}
\mathbf{B} & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{C} & =5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\mathbf{B C} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}
\end{aligned}\)
\(\max \{A B, B C, A C\}=B C\)
\(\therefore \mathbf{B C}\) is hypotenuse of \(\triangle A B C\)
\(\begin{aligned} & \angle \mathbf{A}=90^{\circ} \\ & \therefore \quad \mathbf{A B} \cdot \mathbf{A C}=0 \\ & \mathbf{B A} \cdot \mathbf{B C}=|\mathbf{B A} \| \mathbf{B C}| \cos B \\ & \mathbf{C A} \cdot \mathbf{C B}=|\mathbf{C A} \| \mathbf{C B}| \cos C \\ & \therefore \mathbf{A B}-\mathbf{A C}+\mathbf{B A}-\mathbf{B C}+\mathbf{C A}-\mathbf{C B} \\ &=0+|\mathbf{B C}|(|\mathbf{B A}| \cos B+|\mathbf{C B}| \cos C) \\ &=0+|\mathbf{B C} \| \mathbf{B C}| \quad[\because \text { By projection formula }] \\ &=|\mathbf{B C}|^2-\left(\sqrt{(2)^2+3^2+4^2}\right)^2=4+9+16=29\end{aligned}\)
\(\begin{aligned}
\mathbf{B} & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{C} & =5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\mathbf{B C} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}
\end{aligned}\)
\(\max \{A B, B C, A C\}=B C\)
\(\therefore \mathbf{B C}\) is hypotenuse of \(\triangle A B C\)
\(\begin{aligned} & \angle \mathbf{A}=90^{\circ} \\ & \therefore \quad \mathbf{A B} \cdot \mathbf{A C}=0 \\ & \mathbf{B A} \cdot \mathbf{B C}=|\mathbf{B A} \| \mathbf{B C}| \cos B \\ & \mathbf{C A} \cdot \mathbf{C B}=|\mathbf{C A} \| \mathbf{C B}| \cos C \\ & \therefore \mathbf{A B}-\mathbf{A C}+\mathbf{B A}-\mathbf{B C}+\mathbf{C A}-\mathbf{C B} \\ &=0+|\mathbf{B C}|(|\mathbf{B A}| \cos B+|\mathbf{C B}| \cos C) \\ &=0+|\mathbf{B C} \| \mathbf{B C}| \quad[\because \text { By projection formula }] \\ &=|\mathbf{B C}|^2-\left(\sqrt{(2)^2+3^2+4^2}\right)^2=4+9+16=29\end{aligned}\)
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