Let vecmathbfa = hatmathbfi + 2hatmathbfj + 3hatmathbfk$\vec{\mathbf{a}} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$ , vecmathbfb = 3hatmathbfi + hatmathbfj - hatmathbfk$\vec{\mathbf{b}} = 3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$ and \|\vec{\mathbf{c}}\| be three vectors such that vecmathbfc$\vec{\mathbf{c}}$ is coplanar with vecmathbfa$\vec{\mathbf{a}}$ and vecmathbfb$\vec{\mathbf{b}}$ . If the vector vecmathbfc$\vec{\mathbf{c}}$ is perpendicular to vecmathbfb$\vec{\mathbf{b}}$ and vecmathbfa cdot vecmathbfc = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$ , then \|vecmathbfc\|$\|\vec{\mathbf{c}}\|$ is equal to :

### Related Formula A vector vecmathbfc$\vec{\mathbf{c}}$ coplanar with vecmathbfa$\vec{\mathbf{a}}$ and vecmathbfb$\vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$ can be expressed using the vector triple product command template: vecmathbfc = lambda (vecmathbfb times (vecmathbfa times vecmathbfb)) = lambda [(vecmathbfb cdot vecmathbfb)vecmathbfa - (vecmathbfa cdot vecmathbfb)vecmathbfb]$$\vec{\mathbf{c}} = \lambda (\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})) = \lambda [(\vec{\mathbf{b}} \cdot \vec{\mathbf{b}})\vec{\mathbf{a}} - (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})\vec{\mathbf{b}}]$$ ### Core Logic First, find the dot products of the given vectors: vecmathbfb cdot vecmathbfb = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$\vec{\mathbf{b}} \cdot \vec{\mathbf{b}} = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$ vecmathbfa cdot vecmathbfb = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$ Substituting these values into the expression for vecmathbfc$\vec{\mathbf{c}}$: vecmathbfc = lambda [11vecmathbfa - 2vecmathbfb]$$\vec{\mathbf{c}} = \lambda [11\vec{\mathbf{a}} - 2\vec{\mathbf{b}}]$$ vecmathbfc = lambda [11(hatmathbfi + 2hatmathbfj + 3hatmathbfk) - 2(3hatmathbfi + hatmathbfj - hatmathbfk)]$$\vec{\mathbf{c}} = \lambda [11(\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) - 2(3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}})]$$ vecmathbfc = lambda (5hatmathbfi + 20hatmathbfj + 35hatmathbfk) = 5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)$$\vec{\mathbf{c}} = \lambda (5\hat{\mathbf{i}} + 20\hat{\mathbf{j}} + 35\hat{\mathbf{k}}) = 5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$ ### Step 1: Determine Lambda Using the given condition vecmathbfa cdot vecmathbfc = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$: vecmathbfa cdot [5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)] = 5$$\vec{\mathbf{a}} \cdot [5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})] = 5$$ 5lambda (1 cdot 1 + 2 cdot 4 + 3 cdot 7) = 5$$5\lambda (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 7) = 5$$ 5lambda (1 + 8 + 21) = 5 implies 30lambda = 1 implies lambda = frac130$$5\lambda (1 + 8 + 21) = 5 \implies 30\lambda = 1 \implies \lambda = \frac{1}{30}$$ Thus, the vector vecmathbfc$\vec{\mathbf{c}}$ is: vecmathbfc = frac530(hatmathbfi + 4hatmathbfj + 7hatmathbfk) = frac16(hatmathbfi + 4hatmathbfj + 7hatmathbfk)$$\vec{\mathbf{c}} = \frac{5}{30}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) = \frac{1}{6}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$ ### Step 2: Calculate Magnitude The magnitude of vecmathbfc$\vec{\mathbf{c}}$ is evaluated as: \|vecmathbfc\| = fracsqrt1^2 + 4^2 + 7^26 = fracsqrt1 + 16 + 496 = fracsqrt666 = sqrtfrac6636 = sqrtfrac116$$\|\vec{\mathbf{c}}\| = \frac{\sqrt{1^2 + 4^2 + 7^2}}{6} = \frac{\sqrt{1 + 16 + 49}}{6} = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$$ ### Pattern Recognition Whenever a vector is specified to be coplanar with vecmathbfa, vecmathbfb$\vec{\mathbf{a}}, \vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$, direct setup with the cross-product template vecmathbfb times (vecmathbfa times vecmathbfb)$\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})$ circumvents solving cumbersome linear systems of scalar variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra