SATHEE: Vectors 1 Question 10

Vectors 1 Question 10

11. Let, $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + \hat{j} - \hat{k}$ .A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{\sqrt{3}}$ , then the vector is

(2006, 3M)

(a) $4 \hat{i} - \hat{j} + 4 \hat{k}$

(b) $4 \hat{i} + \hat{j} - 4 \hat{k}$

(d) None of these

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Answer:

Correct Answer: 11. (a)

Solution:

Let vector $\vec{r}$ be coplanar to $\vec{a}$ and $\vec{b}$ .

$\begin{array}{ll} ∴ & \vec{r} = \vec{a} + t \vec{b} \\ \Rightarrow & \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + t (\hat{i} - \hat{j} + \hat{k}) \end{array}$

$= \hat{i} (1 + t) + \hat{j} (2 - t) + \hat{k} (1 + t)$

The projection of $\vec{r}$ on $\vec{c} = \frac{1}{\sqrt{3}}$ .

[given]

$\begin{aligned} \Rightarrow \frac{\vec{r} \cdot \vec{c}}{| \vec{c} |} = \frac{1}{\sqrt{3}} \\ \Rightarrow \frac{| 1 \cdot (1 + t) + 1 \cdot (2 - t) - 1 \cdot (1 + t) |}{\sqrt{3}} = \frac{1}{\sqrt{3}} \\ \Rightarrow (2 - t) = \pm 1 \Rightarrow t = 1 or 3 \end{aligned}$

When, $t = 1$ , we have $\vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$

When, $t = 3$ , we have $\vec{r} = 4 \hat{i} - \hat{j} + 4 \hat{k}$

Vectors 1 Question 10

11. Let, $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + \hat{j} - \hat{k}$ .A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{\sqrt{3}}$ , then the vector is

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Vectors 1 Question 10

11. Let, a→=i^+2j^+k^,b→=i^−j^+k^,c→=i^+j^−k^.A vector coplanar to a→ and b→ has a projection along c→ of magnitude 13, then the vector is

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11. Let, $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + \hat{j} - \hat{k}$ .A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{\sqrt{3}}$ , then the vector is