If \overrightarrow a and \overrightarrow b are vectors in space given by \overrightarrow a = \widehat i - 2\widehat j \over \sqrt 5 and \overrightarrow b = 2\widehat i + \widehat j + 3\widehat k \over \sqrt 14 , then find the value of \,( 2\overrightarrow a + \overrightarrow b ).[ ( \overrightarrow a \times \overrightarrow b ) \times ( \overrightarrow a - 2\overrightarrow b ) ].

Vector Algebra PYQ — JEE Advanced 2010 · Shift 1 · Q42 with Solution

View written solutionFree

Correct answer: 5

We need to evaluate

(2\vec a+\vec b)\cdot\left[(\vec a\times \vec b)\times(\vec a-2\vec b)\right].

Given

\vec a=\frac{\hat i-2\hat j}{\sqrt5}, \qquad \vec b=\frac{2\hat i+\hat j+3\hat k}{\sqrt{14}}.

Use the vector triple product identity:

(\vec p\times \vec q)\times \vec r=\vec q(\vec p\cdot \vec r)-\vec p(\vec q\cdot \vec r).

Here, let

\vec p=\vec a,\quad \vec q=\vec b,\quad \vec r=\vec a-2\vec b.

So,

(\vec a\times \vec b)\times(\vec a-2\vec b) =\vec b\,[\vec a\cdot(\vec a-2\vec b)]-\vec a\,[\vec b\cdot(\vec a-2\vec b)].

First compute the needed dot products.

Since $\vec a$ and $\vec b$ are normalized:

|\vec a|=1,\qquad |\vec b|=1.

Thus,

\vec a\cdot\vec a=1, \qquad \vec b\cdot\vec b=1.

Now,

\vec a\cdot\vec b =\frac{(1)(2)+(-2)(1)+(0)(3)}{\sqrt5\sqrt{14}} =\frac{2-2}{\sqrt{70}}=0.

So $\vec a\perp \vec b$ .

Therefore,

\vec a\cdot(\vec a-2\vec b)=\vec a\cdot\vec a-2\vec a\cdot\vec b=1-0=1,

and

\vec b\cdot(\vec a-2\vec b)=\vec b\cdot\vec a-2\vec b\cdot\vec b=0-2=-2.

Hence,

(\vec a\times \vec b)\times(\vec a-2\vec b) =\vec b(1)-\vec a(-2)=\vec b+2\vec a.

Now the required expression becomes

(2\vec a+\vec b)\cdot(\vec b+2\vec a).

Since this is the same vector,

(2\vec a+\vec b)\cdot(\vec b+2\vec a)=|2\vec a+\vec b|^2.

Expand:

|2\vec a+\vec b|^2=(2\vec a+\vec b)\cdot(2\vec a+\vec b) =4\vec a\cdot\vec a+\vec b\cdot\vec b+4\vec a\cdot\vec b.

Using

\vec a\cdot\vec a=1,\quad \vec b\cdot\vec b=1,\quad \vec a\cdot\vec b=0,

we get

|2\vec a+\vec b|^2=4(1)+1+4(0)=5.

Final answer:

\boxed{5}

Comparison with stored correct answer: stored answer is $5$ , which matches.

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View written solutionFree

Correct answer: 5

We need to evaluate

(2\vec a+\vec b)\cdot\left[(\vec a\times \vec b)\times(\vec a-2\vec b)\right].

Given

\vec a=\frac{\hat i-2\hat j}{\sqrt5}, \qquad \vec b=\frac{2\hat i+\hat j+3\hat k}{\sqrt{14}}.

Use the vector triple product identity:

(\vec p\times \vec q)\times \vec r=\vec q(\vec p\cdot \vec r)-\vec p(\vec q\cdot \vec r).

Here, let

\vec p=\vec a,\quad \vec q=\vec b,\quad \vec r=\vec a-2\vec b.

So,

(\vec a\times \vec b)\times(\vec a-2\vec b) =\vec b\,[\vec a\cdot(\vec a-2\vec b)]-\vec a\,[\vec b\cdot(\vec a-2\vec b)].

First compute the needed dot products.

Since $\vec a$ and $\vec b$ are normalized:

|\vec a|=1,\qquad |\vec b|=1.

Thus,

\vec a\cdot\vec a=1, \qquad \vec b\cdot\vec b=1.

Now,

\vec a\cdot\vec b =\frac{(1)(2)+(-2)(1)+(0)(3)}{\sqrt5\sqrt{14}} =\frac{2-2}{\sqrt{70}}=0.

So $\vec a\perp \vec b$ .

Therefore,

\vec a\cdot(\vec a-2\vec b)=\vec a\cdot\vec a-2\vec a\cdot\vec b=1-0=1,

and

\vec b\cdot(\vec a-2\vec b)=\vec b\cdot\vec a-2\vec b\cdot\vec b=0-2=-2.

Hence,

(\vec a\times \vec b)\times(\vec a-2\vec b) =\vec b(1)-\vec a(-2)=\vec b+2\vec a.

Now the required expression becomes

(2\vec a+\vec b)\cdot(\vec b+2\vec a).

Since this is the same vector,

(2\vec a+\vec b)\cdot(\vec b+2\vec a)=|2\vec a+\vec b|^2.

Expand:

|2\vec a+\vec b|^2=(2\vec a+\vec b)\cdot(2\vec a+\vec b) =4\vec a\cdot\vec a+\vec b\cdot\vec b+4\vec a\cdot\vec b.

Using

\vec a\cdot\vec a=1,\quad \vec b\cdot\vec b=1,\quad \vec a\cdot\vec b=0,

we get

|2\vec a+\vec b|^2=4(1)+1+4(0)=5.

Final answer:

\boxed{5}

Comparison with stored correct answer: stored answer is $5$ , which matches.

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