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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 7 · Set 1 · Q31
(a) The scalar product of the vector a = i - j +2 k with a unit vector along sum of vectors b =2 i -4 j +5 k and c =λ i -2 j -3 k is equal to 1. Find…
(a)
The scalar product of the vector $\displaystyle \vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ with a unit vector along sum of vectors $\displaystyle \vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\displaystyle \vec{c}=\lambda \hat{i}-2 \hat{j}-3 \hat{k}$ is equal to $\displaystyle 1$ . Find the value of $\displaystyle \lambda$.
OR
(b)
Find the shortest distance between the lines: \[\begin{aligned} & \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(\hat{i}-2 \hat{j}+3 \hat{k}) \\ & \vec{r}=(\hat{i}+4 \hat{k})+\mu(3 \hat{i}-6 \hat{j}+9 \hat{k}) \end{aligned} \]
Marking-scheme solution
(a)
Let $\displaystyle \overrightarrow{\mathbf{d}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=(\mathbf{2}+\boldsymbol{\lambda}) \hat{\mathbf{i}}-\mathbf{6} \hat{\mathbf{j}}+\mathbf{2} \hat{\mathbf{k}}$
\[\hat{\mathbf{d}}=\frac{(2+\lambda) \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{(2+\lambda)^{2}+40}}
\]
OR
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